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2) A professor is studying exam scores. a) State the Hypothesis to show the vari

ID: 3182040 • Letter: 2

Question

2) A professor is studying exam scores.

a) State the Hypothesis to show the variance in exam score is different than 64.

b) Choose a level of Use = 0.05 for this problem.

c) To test the hypothesis, the professor takes a sample of 25 exams and records the score. The data appear in the Exams worksheet of the HW11 data workbook on Moodle. Collect data and calculate necessary statistics to test the hypothesis.

d) Sketch the sampling distribution. Include the critical value and test statistic

e) Draw a conclusion and report that in the problem context.

f) Calculate the p-value for the hypothesis test

Score 80 73 81 78 88 79 77 84 62 70 69 75 84 77 75 71 65 79 68 85 74 52 70 77 84

Explanation / Answer

A) H = the exam score variance = 64

H = the exam score variance = 64

b) = 0.05

confidence interval = 95%

c)

1877

count 25

these values are calculated in excel, you can do it manually too.you will need only variance of the sample as test statistics for hypothesis testing.

d) t -test

t= - x/ 8.11/25-1

= 64 - 65.82/8.11/24

=- 5.38

d)conclusion= 95% values of variance of the population will fall into 64-5.38 = 58.62 and 64+5.83 =69.83. these are the critical values which shows that 95% of the values of sample will fall in this range.

this a small sample where the chance of error in estmination increases. to have a good estimation of test statistic the no. of obeservation(exam scores in this case) should be greater than30.

e) p value is -5.38(by looking at table) it comes undeer 90% confidence interval which shows that the population mean is significantly diffrent from 64. thus the null hyposthese is rejected.

Mean 75.08 Standard Error 1.622672692 Median 77 Mode 77 Standard Deviation 8.113363462 Sample Variance 65.82666667 Kurtosis 1.375179255 Skewness -0.909160368 Range 36 Minimum 52 Maximum 88 Sum

1877

count 25

these values are calculated in excel, you can do it manually too.you will need only variance of the sample as test statistics for hypothesis testing.

d) t -test

t= - x/ 8.11/25-1

= 64 - 65.82/8.11/24

=- 5.38

d)conclusion= 95% values of variance of the population will fall into 64-5.38 = 58.62 and 64+5.83 =69.83. these are the critical values which shows that 95% of the values of sample will fall in this range.

this a small sample where the chance of error in estmination increases. to have a good estimation of test statistic the no. of obeservation(exam scores in this case) should be greater than30.

e) p value is -5.38(by looking at table) it comes undeer 90% confidence interval which shows that the population mean is significantly diffrent from 64. thus the null hyposthese is rejected.

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