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The admissions officer for the graduate programs at the University of Adelaide believes that the average score on an exam at his university is significantly higher than the national average. Assume that the population standard deviation is 125 and that a random sample of 25 scores had an average of 1375.

1. State the appropriate null and alternative hypotheses.
2. Calculate the value of the test statistic and set up the rejection region. What is your conclusion?

c.Calculate the p- value

d.Does the p-value confirm the conclusion in part (b)?

i)

A statistician wants to estimate the mean weekly family expenditure on clothes. He believes that the largest weekly expenditure is \$650 and the lowest is \$150.

a.Estimate the standard deviation of the weekly expenditure.

b.Determine with 99% confidence the number of families that must be sampled to estimate the mean weekly family expenditure on clothes to within \$15.

ii)

The head of the statistics department in a certain university believes that 70% of the department’s graduate assistantships are given to international students. A random sample of 50 graduate assistants is taken.

a.Assume that the chairman is correct and p = 0.70. What is the sampling distribution of the sample proportion? Explain.

b.Find the expected value and the standard error of the sampling distribution of?

c.What is the probability that the sample proportion will be between 0.65 and 0.73?

d.What is the probability that the sample proportion  will be within ±0.05 of the population proportion p?

## Problem 1: Exam Scores and P-values

a)

Based on the regression output we have to find out the trend equation and interpret. Now, from the output we get that it is an output of a simple linear regression i.e. there is only one response variable and only one independent variable. Here the response variable is goggle sales(in thousands of dollars) denoted by “y” and the independent variable is time variable denoted by “t”. Here the origin of the data is the march quarter 2000.

Here the coefficient of intercept is 12.237 and the coefficient of the time variable is 0.26289.

So the trend line is, y(in thousands of dollars)=12.237+0.26289 t   (Origin: March Quarter 2000)

Interpretation: Hence for each unit increase in the time variable t i.e for each quarter from march quarter 2000 there will be an increase of 0.26289 unit in the goggle sales (in thousands of dollars).

b)

95% Confidence interval of the variable is given by,(0.080812368,0.4449738)

According to the confidence interval if we collect samples again and again from the population again and again then 95% of the times the coefficient of time variable t will lie within this interval i.e for each unit of increase in the time variable t there will be an increase in the response variable i.e in the goggle sales (in thousands of dollars) will lie within the interval (0.080812368,0.4449738)

c)

Management report about forecast of goggle sales of each quarter of 2016

Introduction:

According to the researcher’s problem a company which is selling swimming goggles desires to investigate the company’s Australian sales. Dataset of 54 observations were collected and based on that a time series forecasting model using Regression technique was used. Our main aim is to forecast sales of goggle sale in the upcoming years.

Here, the data was used to generate Excel output using the Excel Data analysis function to generate the summary output and from the summary output we are able to estimate trend of the time series of Swimming goggle sales. Here the unit of sales taken is in thousands of dollars.

We have made the origin of the data as March Quarter 2000. In this way we have changed the

data “t” in such a way that the origin of  “t” is in March Quarter 2000. So, the data observed in 2000’s second quarter is t=1,third quarter t=2… and so on.

Our motivation is to find out how the sales of swimming goggle depend in the time variable i.e. the degree to which the sales are predicted by the year. Another aim is to forecast or predict the number of sales in the upcoming years. This is done by many companies so that they could know how much products to be manufactured so that they are on the profit side.

## Problem 2: Family Expenditure and Confidence Intervals

At first we have found out the correlation i.e. the association among the two variables . Then we have done a regression analysis along with its anova table is given from which we have got full information about the data which helps us to predict the goggle sales in the future quarters.

We have used regression analysis and one way anova . Here our dependent variable is Goggle sales and independent variable is Year. We have here considered only simple linear regression.

Output and Interpretations: From the data, Using Excel we get the following results using “Data Analysis” Tool in Excel.

The following table gives the regression parameters’ coefficients, t statistic value and p value associated with them . Also the 95% upper and lower confidence intervals are also given.

 Coefficient Standard Error T statistic P value Lower 95% Upper 95% intercept 12.237 2.7896 4.39 5.6E-05 6.64 17.83 t 0.262 0.0907 2.897 0.0055 0.0808 0.445

Formula:

Independent Variable: X

Output Variable: Y

Number of Observation: n

Correlation Coefficient = = R

Coefficient of Determination=

T statistic =

Table 1: Table showing Regression Parameters and their properties of Google sales vs Year

Here, Multiple correlation coefficient R=0.37281

R(square) or Coefficient of Determination= 0.13899

Standard Error=10.3925

 Df Sum of Squares Mean Sum of Squares F statistic Significnce F Regression 1 906.5867925 906.59 8.39406 0.005497292 Residual 52 5616.172467 108 - - Total 53 6522.759259 - - -

Table 2: Table showing Anova table associated with the regression of Google sales vs Year

Interpretation of  Correlation Coefficient:

Here, the correlation coefficient is given as 0.37281. Hence there is a positive trend in the data as

time increases the goggle sales will increase. It implies the company is running in profits.

Interpretation of standard Error:

Here in the regression output we get the standard error which is associated with the regression

problem. This can be used to evaluate the accuracy of the forecasts. Standard error is used mainly to compute the accuracy of the forecasts as with the help of it we can get the limit in which the 95% of the values should lie . It gives the interpretation that they should lie inside  2* Standard Error of the regression. In this way we can quickly deduce an estimate of the  prediction interval precisely 95%. As the standard error is 10.3925 so the prediction interval will be 2*10.3925

From the above table we get  the coefficients of intercept and t variable. P values associated with

them is less than 0.05(significance level) which implies the regression is significant.

Interpretation of the confidence intervals of the coefficients of “t” and intercept:

As the confidence interval of intercept is (6.64,17.83) i.e. it does not contain 0 so it is significant.

## Problem 3: Sample Proportions and Probability

On the other hand confidence interval of coefficient of “t” is (0.0808,0.445) so it does not contain

0 hence it is also significant.

Adjusted R square take into account the number of predictors. It penalizes for adding number of predictors.

Here after collection of data and analyzing we get the following linear trend equation as, y(in thousands of dollars)=12.237+0.26289 t   (Origin: March Quarter 2000)

The response variable is goggle sales(in thousands of dollars) denoted by “y”

The independent variable is time variable denoted by “t“.

Note that

1) Here we have the value of coefficient of determination( R-square)=0.13899. Hence out of the total variability in the response variable only 13.9% is explained by our linear trend equation.

2) Our trend equation is based on 54 observations.

3) From the regression output and the p values of the coefficients the regression coefficients are significant (as P value<0.05 i.e. the level of significance hence we reject the null hypothesis i.e  . , i=1,2  are  the coefficients of the intercept and the time variable respectively)

EstimatesForecasts of goggle sales of each quarter of 2016

In each year there are four quarters. Here the four quarters are Jan-March, April-June, July-September, October-December.

For the first quarter of 2016, t=64 . Hence the estimated goggle sales of first quarter of 2016 is give by. Y= 12.237+0.26289 *64 =29.06196 (in thousands of dollars)

For the second quarter of 2016, t=65. Hence the estimated goggle sales of second quarter of 2016 is give by. Y= 12.237+0.26289 *65 =29.32485 (in thousands of dollars)

For the third quarter of 2016, t=66. Hence the estimated goggle sales of third quarter of 2016 is give by. Y= 12.237+0.26289 *66 =29.58774 (in thousands of dollars)

For the fourth quarter of 2016, t=67. Hence the estimated goggle sales of fourth quarter of 2016 is give by. Y= 12.237+0.26289 *64 =29.85063 (in thousands of dollars).

The difference between each quarter’s goggle sales is 0.26289 and there is an increasing trend in the goggle sales i.e. goggle sales are increasing over the time which is evident as the coefficient of the regression is positive.

The sales of goggle helps the company to know how much to produce . If we had obtained more data we could obtain a better prediction. Also here the coefficient of determination is 0.13899 i.e. only 14% of the total variation of the sales is explained by the time variable. Hence we should check higher degree of regression i.e. we should check whether the data follows a quadratic trend or cubic trend . We could have an idea of the trend it is following by a scatterplot . But above all we should take more values into consideration so that our output result is better.

Question 3

1. a) According to the problem the suitable null and alternative hypotheses are given as below, :μ=1300      : μ>1300     (μ is denoted as the population mean of the exam at the university)

Here the appropriate test statistic is given by,

T= ~ N(0,1) (Under the null hypothesis)

Here, μ=1300 and  is the sample mean scores on the exam at the university and  is the population standard deviation of the marks of the exam at the university, n denote the sample size.

Here  = 125 , =1375 ,n=25

Putting the values we get the value of the statistic as, obs(T)=3

Here we take the level of significance as α=0.05 .

Hence we reject the null hypothesis if obs(T)>  .

is the 100* α % point of standard normal distribution.

Here =1.64485 (for α=0.05 )

Hence obs(T)>   . So at 5% level of significance in the light of the data, there is enough evidence to support the claim that the average score on an exam at the university which is under concern is significantly higher than the national average of 1300

P value= P(T>obs(T))=P(T>3)= 0.00135    (Here T follows Standard Normal distribution)

Yes the  P value confirms the conclusion in part (b) because here the p value< α i.e the level of significance. So we reject the null hypothesis and support the claim that the average score on an exam at the university is significantly higher than the national average of 1300. So we get the same conclusion as in (b)

1. ii)H0: μ= 50   ,  HA: μ > 50

given that  µ = 55, α = 0.05, s = 10 and n = 16.

Type II error is accepting a false null hypothesis.

Here we accept the null hypothesis if the obs(T)<  (α = 0.05)  where is the 100* α % point of standard normal distribution.

=1.64485 (α = 0.05)

Hence we have to find the type II error means we have to find the probability that the observed test statistic value lies under  when the null hypothesis is false i.e when µ = 55 .

Z, follows Standard Normal distribution.

P(  )=P(<+ 50) = P()=P(Z<-0.35515)= 0.361239

Hence the probability of Type II error 0.361239 .

References:

Sheldon, Ross (2010). Introductory Statistics, Academic Press,USA.

Hoel,P.G.,(1971),Introduction to Mathematical Statistics,Fourth Edition,USA

Feller,William(2013),An introduction to Probability Theory and Its Applications,Volume I,Third Edition,U.K.

Du Prel, J.-B., Hommel, G., Röhrig, B., & Blettner, M. (2009). Confidence Interval or P-Value?: Part 4 of a Series on Evaluation of Scientific Publications. Deutsches Ärzteblatt International, 106(19), 335–339.

Akobeng AK. Confidence intervals and p-values in clinical decision making. Acta Paediatr. 2008;97:1004–1007

Altman D, Bland JM. Confidence intervals illuminate absence of evidence. BMJ. 2004;328:1016–1017

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