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Sample mean, = 45; sample standard deviation, s = 10; sample size, n = 70
The formula for the confidence interval of population mean is defined as,
Where, = = 1.1952
z = 2.58 (obtained from the z table for α/2 = 0.005)
Margin of error = = 3.0837
Thus, lower limit of the confidence interval = 45 – 3.0837 = 41.9163 and the upper limit = 45 + 3.0837 = 48.0837.
Therefore, 99% C.I. is (41.9163, 48.0837).
The formula for the confidence interval of population proportion estimate is defined as,
The level of significance = α = 1 95 = 0.05.
The value of critical z at α/2 = 0.05/2 = 0.025 is obtained from the z table as 1.96.
Standard error of = = 0.059737.
Margin of error = 1.96 0.059737 ≈ 0.117085.
The lower limit of the 95% confidence interval = – 0.117085 = 0.3972 and the upper limit of the confidence interval =.
Therefore, 95% C.I. is (0.3972, 0.63137)
Sample mean sample size, n = 70, sample standard deviation, s = 9.22.
The formula of 95% confidence interval for the population mean is given by,
t_{(}_{α}_{/2, n1)} = 21.34 ( = 21.34
= 21.34
Thus, the lower limit of the confidence interval = 21.34 – 2.198859 = 19.14114 and the upper limit = 21.34 + 2.198859 = 23.53886.
Therefore, the confidence interval of mean amount spent in the pet supply store is (19.14114, 23.53886).
= = = = 0.371429.
The significance level, α = 1 – 90 = 0.10.
From the standard normal table, Z_{α/2 }= Z_{0.10/2 }= Z_{0.05} = 1.645. The formula of confidence interval is
Z_{0.05} = 1.645 = 1.645 = 0.095002
The confidence interval is 0.371429 0.095002 that is, (0.276427, 0.46643).
From the ztable, the confidence coefficient Z_{α/2 }= Z_{0.025 }= 1.96.
The sample size (n) is calculated as n = = = = 170.7378.
Therefore, the sample size is 170.
The given information are proportion of population, π = 0.5, margin of error, e = 0.045. The level of significance, α = 10.90 = 0.10.
From the standard normal table, Z_{α/2 }= Z_{0.10/2 }= 1.645.
Thus, the required sample size, n = = = = 334.0772.
Therefore, the sample size is 335.
The given information are sample mean, = 5.5, standard deviation, s = 0.099, sample size, n = 50. The standard error, e = s/ = 0.099/50 = 0.014.
The critical value of z at α = 0.01 is 2.58 (from the ztable).
The confidence interval is given by 2.58.
The margin of error = 2.58 = 0.0362.
Therefore, the 99% confidence interval for the population mean weight is given by 5.5 that is, (5.4638,5.5362).
The sample mean is calculated from the given data of 50 observation and it is 5.497.
It is clearly seen that the sample mean lies within the confidence interval. Thus, the null hypothesis is accepted and it can be concluded that the company is meeting the requirement of containing 5.5 grams tea in the tea bag.
H_{0}: The proportion of population of web page visitors preferring the new design is 0.50.
H_{1}: The proportion of population of web page visitors preferring the new design is not 0.50
In the given course of problem, when the population proportion of web page visitors preferring the new design is truly 0.50 but the investigator reject the null hypothesis, then type I error occurs. Again, if the investigator accepts the null hypothesis when the population proportion of web page visitors preferring the new design is actually not 0.50, then the type II error occurs.
If the pvalue is reduced to 0.06, then also the null hypothesis will not be rejected as the pvalue is still greater than the significance level.
The null hypothesis for the required test is given by
H_{0}: µ 100 against the alternative hypothesis H_{1}: µ < 100 where µ denotes the population mean.
The test statistic, t_{cal} = () = 1.58 with degrees of freedom 751 = 74 and the significance level = 0.05
The critical value of tstatistic is 1.6657 which is greater than 0.05. Thus, the null hypothesis is accepted and there is not enough evidence to state that the mean reimbursement is less than $100.
H_{0}: π 0.10 against H_{1}: π > 0.10; where π is the population proportion.
Sample proportion p = 12/75 = 0.16
The calculated value of the teststatistic of this onetail ztest = = 1.73.
The null hypothesis is rejected as 1.73> 1.64, where 1.64 is the critical zvalue at 5% level of significance.
Clearly, 2.51< 1.64, the null hypothesis is rejected at 5% level of significance and the mean reimbursement value will be less than $100.
At 5% significance level, the H_{0} will be rejected here as 2.89>1.64. Thus, there is enough proof to have the population proportion more than 0.10.
The required calculations are shown in the following table
Measurements 
Gender (Female) 
Gender (Male) 

18 
10 

20 
5.5 

10 
6 

18 
14 

8 
10 

25 
6 

22.5 
12 

30 
21 

20 
15 

24 
12 

15 
10 

15 
8.5 

10 
25 

14 
10 

15 
20 

12 
8 

11 
8 

12 
12 

15 
6 

18 
6 


7 


10 


10 


9 


10 


30 


10 


10 


5 


5 


10 


11 


15 


15 


7 


8 


12 


10 
Count 
20 
38 
Mean 
16.625 
11.0263158 
Standard Deviation (SD) 
5.71442082 
5.40638682 
The population of females is represented by population 1 and that of males is represented by population 2.
Thus, n_{1} = 20, n_{2} = 38, = 16.625, = 11.0263, s_{1} = 5.7144, s_{2} = 5.4064.
H_{0}: = against H_{1}: ≠ ; where and represents the population variances of population 1 and population 2 respectively.
The test statistics is given by,
F = = 1.117.
Under null hypothesis, the test statistics follow F_{201, 381} = F_{19, 37} at 5% level of significance.
The pvalue is given by 0.7492 which shows that the pvalues is greater than 0.05 and therefore the null hypothesis is accepted. Thus, it can be concluded that the population variances are equal for both the populations.
The pooled standard deviation is calculate3d as,
S_{p }= = 5.5128.
And tstatistics is calculated as t = = 3.676 with degrees of freedom = 20+382 = 56.
The present test is twotailed and the pvalue is 0.0005 which shows that the pvalue is less than the significance level 0.05. Therefore, the null hypothesis is rejected and finally the mean time for study of the two populations are not equal.
The populations of Pinterest shoppers and the Facebook shoppers be respectively denoted as population 1 and population 2.
Given that, n_{1} = 500, n_{2} = 500, = 153, = 85, s_{1} = 150, s_{2} = 80.
The test statistics is given by,
F = = = 3.5156.
At 5% level of significance, the critical value of F with (499,499) degrees of freedom is 1.16 (obtained from the Ftable).
Thus, the observed value of F is greater than the critical value of F which leads to reject the null hypothesis. Therefore there is difference in the variance of the order values of Pinterest shoppers and the Facebook shoppers.
The pooled variance is S^{2} = = 12526.15
The test statistic is t = = 9.61.
The critical value of t statistic with df = 500+5002 = 998 at 5% level of significance is given as 1.96.
The critical value is less than the calculated value of tstatistic. Thus, the null hypothesis is rejected and finally the mean time for study of the two populations are not equal.
µ_{1}  µ_{2} =  ± _{ } = (15385) ± 1.96 _{ }
_{ } Thus, the 95% confidence interval is given by (53.09, 82.90).
Therefore, the simple linear regression equation is given by:
Where is the estimated value of Y corresponding to the values of X, is the yintercept of the sample and is the sample slope. The values of i run from 1 to 7.
The Data Analysis Tool Pak of MS Excel is used to calculate the simple linear regression line.
First, the data is written in the Excel sheet. Then, click on the “Data” tab à Click on the “Data Analysis” tab
à Choose regression à select thee option and the data region as shown in the image belowàclick OK.
The output is obtained as,

Coefficients 
Standard Error 
t Stat 
Pvalue 
Lower 95% 
Upper 95% 
Lower 95.0% 
Upper 95.0% 
Intercept 
6808.104703 
854.9682277 
7.962991468 
0.00050374 
4610.338906 
9005.8705 
4610.3389 
9005.8705 
Twitter Activity 
0.050271541 
0.003527041 
14.25317587 
3.0627E05 
0.041204992 
0.0593381 
0.041205 
0.0593381 
Thus, intercept b_{0} = 6808.1047 and the slope b_{1} = 0.0502715.
Therefore, the predicted value of the receipts fir a movie that ha TWITTER ACTIVITY OF 100,000 is given as = $11835.259.
Regression Statistics 

Multiple R 
0.987916601 
R Square 
0.97597921 
Adjusted R Square 
0.971175052 
Standard Error 
1830.236093 
Observations 
7 
From the table, it is seen that, the value of r^{2 }≈ 0.9760. This value interprets that, 97.6% variation in the Receipts value is due to the variation in the Twitter activity. The remaining 2.4% variation is due to other factors which are not controlled by the Twitter activity.
RESIDUAL OUTPUT
Twitter Activity 
Receipts 
Residuals 
219509 
14763 
3080.1604 
6405 
5796 
1334.0939 
165128 
15829 
719.65628 
579288 
36871 
941.19487 
6564 
8995 
1856.9129 
11104 
7477 
110.680106 
9152 
8054 
785.810154 
The residual plot is given as,
From the above graph, it is seen that there is no significant pattern in the residual plot.
H_{0}: β_{1} = 0 (There is no linear relationship)
H_{1}: β_{1}≠ 0 (There is a linear relationship)
The ANOVA table is obtained in the Excel sheet which is given below:
ANOVA 






df 
SS 
MS 
F 
Significance F 
Regression 
1 
680514712.65 
680514712.65 
203.15 
0.00 
Residual 
5 
16748820.78 
3349764.16 


Total 
6 
697263533.4 




Coefficients 
Standard Error 
t Stat 
Pvalue 
Lower 95% 
Upper 95% 
Lower 95.0% 
Upper 95.0% 
Intercept 
6808.104703 
854.9682277 
7.962991468 
0.00050374 
4610.338906 
9005.8705 
4610.3389 
9005.8705 
Twitter Activity 
0.050271541 
0.003527041 
14.25317587 
3.0627E05 
0.041204992 
0.0593381 
0.041205 
0.0593381 
From the ANOVA table, it is seen that, the value of tstatistic = 14.2532. Besides, the critical value of tstatistics at 5% level of significance is given as 2.5706 for 72 = 5 degrees of freedom.
Clearly, the calculated value of t –statistic is greater than the critical value which implies that the null hypothesis is rejected. Hence, there is significant linear relationship between Twitter activity and Receipts.
Where 11835.259, 2.5706, S = 1830.2361, = 0.1495492.
Thus, 95% confidence interval estimate of the mean receipts for a movie having a twitter activity is 11835.259 1819.422946 that is (1005.83605, 13654.68195).
Now, the 95% prediction interval of the receipts for a single movie that has a twitter activity of 100,000, is calculated using the following formula:
Therefore, the 95% confidence interval is given as 11835.259 5044.352206 that is, (6790.906794, 16879.61121).
Hence, the yintercept b_{0}= 0.4872 and slope, b_{1}= 0.0123. The equation of linear regression line is Y = 0.04872 + 0.0123X.
The following diagram shows the residual plot against the Time variable:
Here, normal assumption is not specified as there is no significant pattern of the data.
H_{0}: b_{1} = 0 against H_{1}: b_{1} ≠ 0
Test statistic = ~ t_{n2 }
= 13.71
The pvalue=0.00 which fails to accept the H_{0}. Thus, the linear regression model is significant.
The number of wins is the dependent variable and the independent variable is ERA.

Coefficients 
Standard Error 
t Stat 
Pvalue 
Lower 95% 
Upper 95% 
Lower 95.0% 
Upper 95.0% 
Intercept 
168.7804 
14.81386 
11.39341 
5.01E12 
138.4355 
199.1252 
138.4355 
199.1252 
E.R.A.(X) 
22.7194 
3.814214 
5.95652 
2.06E06 
30.5325 
14.9064 
30.5325 
14.9064 
The intercept b_{0}= 168.7804, slope b_{1}= 22.7194. The regression equation is
Wins = 168.7804 – 22.7194 ERA.
The scattered points have less variability around the least square line and there is no outlier. Moreover, the trendline is linear.
The residual plot is shown below:
The residual values are spread equally in the above plot in the middle.
H_{0}: β_{1} = 0 against H_{1}: β_{1} ≠ 0
The pvalue is 0.00 at 5% level of significance which shows that the null hypothesis is rejected and there is linear relationship between the Wins and ERA.
In the MINITAB output above, the 95% CI for the mean wins is (60.6891, 72.3967)
b_{1}= 22.71944336
S_{b1} = 0.1857
From the ttable, t_{(0.025,28)} = 2.0484.
The C.I. is obtained as 22.7194 ± (2.0484*0.1857) that is, (23.0998,22.339).
Thus, the slope of X_{1 }is 1.069 when X_{2 }is 2.
The slope of X_{1} is 0.119.
The slope of X_{2 }is 1.082.
= 3.888 + (1.449) + (1.462 X_{2i}) – (0.19 X_{2i}) = 6.255+0.132 X_{2i }and the slope of X_{2 }is 0.132.
The predicted purchase behavior increases from 1.174 to 6.584 as X_{2 }increases from 2 to 7, remaining the value of X_{1 }at 2 for parts (a) and (b). From (a) and (c), the increase from 1.174 to 6.519. However, in parts (b) and (d), increases to a very small extent. This happens due to negative effect.
The regression equation Wins () = 217.4739 + (4.3522129 Field Goal %) + (1.6921454 ThreePoint Field Goal %)
Pint Field Goal % constant. Again, if the Field Goal % is kept fixed, then one unit increase in the ThreePoint Final Goal will increase Wins by 1.6921454 units.
The errors are normally distributed and the error points are on the trendline.
The residual plot of Field Goal % is shown below that shows there is no patterns and the variances are equal.
The residual plot of Threepoint Field Goal % is displayed below that indicates the points are not equally spread and they are concentrated between 30% and 40%.
Significance level, α = 0.05. The ANOVA table is given below:
ANOVA 






df 
SS 
MS 
F 
Significance F 
Regression 
2 
2397.56149 
1198.7807 
13.273691 
9.6704E05 
Residual 
27 
2438.43851 
90.312537 


Total 
29 
4836 



The large value of F shows that much variation in Wins is explained and as F larger than the critical F, thus the null hypothesis is rejected and the regression model is valid.
The regression model is fitted in MS Excel.
Wins = 87.72133+ (18.9257) ERA + (15.96262) Runs scored per game
= 87.72133 (18.9257 4) + (15.96262 4) = 63.85048 – 75.7028 + 87.72133 = 75.86901 ≈ 76.
The above plot shows that the data is not equally spread and they are concentrated between 3.00 and 4.50. There is no pattern.
In the above residual plot, the points are again concentrated and the points are from 3.00 to 5.00.
Significance level, α = 0.05. The ANOVA table is given below:
ANOVA 






df 
SS 
MS 
F 
Significance F 
Regression 
2 
3828.662 
1914.331 
102.282 
2.51392E13 
Residual 
27 
505.3375 
18.7162 


Total 
29 
4334 



The large value of Fstatistic indicates that the null hypothesis is rejected and the model is valid.
b_{1} ± 1.96
Now, = (0.402273567/ 2.166309458) = 0.185695338
The 95% C.I. is ((18.9257) ± 1.96 0.185695) that is, (19.28966, 18.56174).
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