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Question 1(a)

Suppose that the amount of time teenagers spend on the Internet is normally distributed, with a standard deviation of 1.5 hours. A sample of 100 teenagers is selected at random, and the sample mean is computed as 6.5 hours.

(i) Determine the 99% confidence interval estimate of the population mean.
(ii) Determine the 95% confidence interval estimate of the population mean, changing the sample size to 300.
(iii) Comment on how the confidence interval and changes to sample size will impact on the interval estimates.

Question 1(b)

The amount of time spent by Australian adults playing sports per day is normally distributed, with a mean of 4 hours and standard deviation of 1.25 hours. Use this information to answer the following question(s).

(i) Find the probability that a randomly selected Australian adult plays sport for more than 5 hours per day.
(ii) Find the probability that if four Australian adults are randomly selected, their average number of hours spent playing sport is more than 5 hours per day.
(iii) Find the probability that if four Australian adults are randomly selected, all four play sport for more than 5 hours per day

Question 1(a)

Normal distribution

Standard deviation   = 1.5 hours

Sample mean = 6.5 hours

Sample size = 100

99% confidence interval for the population mean

Z value would be considered as the distribution is normal and sample size is higher than 30.

Z value for 99% confidence interval = 2.58

Confidence interval

Lower limit

Upper limit

99% Confidence interval = [6.113    6.887]

95% confidence interval for the population mean when the sample size is 300

Z value would be considered as the distribution is normal and sample size is higher than 30.

Z value for 95% confidence interval = 1.96

Confidence interval

Lower limit

Upper limit

99% Confidence interval = [6.33    6.67]

The standard error is inversely proportion to the sample size and as sample size increases the standard error would decrease. This is would lead to a decrease in the margin of error which in turn would lead to a narrower confidence interval as with higher value the prediction can be made with higher accuracy.

Mean = 4 hrs

Standard deviation 1.25 hrs

Probability that a randomly selected Australian adult plays a sport for more than 5 hours per day

Therefore, the probability would be 21.19%.

The probability that if four Australia adults are randomly selected, then their average number of hours spent playing sport is more than 5 hours per day

Therefore, the probability would be 43.64%.

The probability that if four Australian adults are randomly selected, then all four play sport for more than 5 hours per day

Hence, the probability would be 0.00201.

The proportion of students that passed her courses = 80%

Sample size = 25

Number of students who have passed the course = 11

Alpha = 0.01

Proportion of students would be = 11/25 = 0.44

Hypothesis testing:

Alternative hypothesis - :

Z statistics would be used.

The p- value for the above computed value of z statistic comes out to be 0.00. Considering that p value is lesser than the level of significance, hence there is sufficient evidence available for the rejection of the null hypothesis. Hence, it may be concluded that proportion of students who would pass the course would significantly deviate from 80%.

Sample size = 50

Sample mean = 2.2 hrs

Sample standard deviation = 3 hrs

Claim -  That the new drug increases the number of hours of sleep at least by 2 hours on average at 5% level of significance.

Hypothesis testing:

Alternative hypothesis - :

As the population standard deviation is not known, hence, t statistics would be used.

The p value for the above t for 49 degree of freedom comes out as 0.3199. Considering that p value is greater than the level of significance (0.05), hence there is insufficient evidence available for the rejection of the null hypothesis. Thus, it may be concluded that the claim for the new drug is not supported by the current sample.

At least two of the proportion is different from given values.

The data has been summarised below:

The observed value () has been shown below:

## Question 1(b)

 Observed values ) Expected value () Brand A 280 300 1.33 Brand B 270 240 3.75 Brand C 90 120 7.50 Brand D 560 540 0.74 Total 1200 13.32

Degree of freedom = 3*1 = 3

p value for Chi square stat 13.32 and 3 degree of freedom is 0.003993.

Level of significance= 1%

It is apparent that p value is lower than the level of significance and therefore, there is sufficient evidence present to reject the null hypothesis. Therefore, it can be said that the market has changed sicne2003.

The two variable gender and insurance are independent variables.

The two variables gender and insurance are dependent variables.

Degree of freedom = (2-1) *(3-1) = 2

p value for Chi square stat 8.4 and 2 degree of freedom is 0.014996.

Level of significance= 5%

It is apparent that p value is lower than the level of significance and therefore, there is sufficient evidence present to reject the null hypothesis. Therefore, it can be said that gender and insurance are dependent variables.

The data has come from a normal probability distribution.

The data has not come from a normal probability distribution.‘

Mean = 26.80

Standard deviation = 6.378

Level of significance = 5%

As the sample size is lower than 80, hence, the minimum number of interval can be decided as four.

and corresponding p value would be 0.0316.

It is apparent that p value is lower than significance level and therefore, sufficient evidence present to reject null hypothesis. Hence, it can be said that data has not come from a normal probability distribution.

Based on the given regression output, the regression model equation may be stated as shown below.

Coffee Sales Revenue (in \$ ’00) = 27.7179 – 0.6943* Temperature (in 0C)

In the given case, temperature = 380 C

Hence, coffee sales revenue (in \$ ’00) = 27.7179 -0.6943*38 = 1.3345

The prediction in part (a) seems to be reasonable considering the fact that given regression model has a coefficient of determination of 0.7472 which implies that temperature alone is a significant variable to explain 75% of the variation in coffee sales revenue. As a result, while the estimate might not be 100% accurate but it would be fairly accurate.

The relevant hypotheses are as stated below.

H0: β = 0 which implies that the slope of the given regression line is not significant and thus can be assumed to be zero.

H1: β ≠ 0 which implies that the slope of the given regression line is significant and thus cannot be assumed to be zero.

From the regression output highlighted in the question, it is apparent that the relevant t statistic for the slope coefficient of temperature comes out as – 3.4387. Further, the corresponding p value comes out as 0.0263.

As per the question, assumed level of significance = 5%

Considering that p value is lesser than the level of significance, hence there is sufficient evidence available for the rejection of the null hypothesis. Hence, it may be concluded that the slope of the regression model is significant.

The regression output and the above findings clearly indicate that there is a strong negative association between the underlying temperature and the revenue generated from coffee vending machine. This is reflected from the fact that R2 or coefficient of determination has a significantly high value. Also, the slope of the regression model is found to be significant. Further, considering this model and the fact that Darwin has a temperature of 38 0C, it is apparent that the sales generated from the coffee vending machine on a daily basis would be quite low. Thus, based on the above analysis, it may be recommended that operating a pop-up coffee machine in Darwin is not a good idea and must be aborted

The objective here is to compare the marks that students tend to obtain in statistics in Semester 1 and Semester 2. In order to enable the same, a sample data has been extended of the marks of 25 students in statistics picked from the two semesters under question. From the sample data, we need to drive the conclusion about the population of statistics marks in the two semesters through the following process.

H0: µsemester 1 = µsemester 2

H1: µsemester 1 ≠ µsemester 2

For the purposes of hypothesis testing, the level of significance has been assumed as 0.05.

The marks distribution for the population of the students is considered to be normal. However, the standard deviation of the population of all students in Semester 1 and Semester 2 in unknown coupled with the sample size being lower than 30. Thus, considering these two conditions, it would be best to deploy t statistic for the purposes of hypothesis testing. There are two options with regards to T test for two independent samples i.e. either the variances could be equal or unequal. In the given case, equal variances would be preferred as the observation count is the same and also the sample variances are not very different as apparent from the given descriptive statistics.

The relevant p value for the given hypothesis testing would be two tail which has the value 0.27. This is apparently greater than significance level of 0.05. It is indicative of the presence of inadequate evidence to bring about null hypothesis rejection. Consequently, there will not be not be acceptance of alternative hypothesis. Considering the non-rejection of null hypothesis, the final conclusion is that the difference between the marks scored by students in statistics for the two semesters is not significant.

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