Question

A random sample of 22 STAT 250 students was collected and the file size of Data Analysis 2 was recorded. The data was measured in megabytes. The instructors of the course claim that the file size will be different from 5 megabytes. Consider the population of all file sizes to be right-skewed. Using a = 0.01, is there sufficient evidence to conclude that the mean file size of Data Analysis 2 is different from 5 megabytes? Conduct a full hypothesis test by following the steps below. Enter an answer for each of these steps in your document.

Data:

Megabytes
7.72
4.49
8.17
5.51
9.09
7.41
6.31
2.04
5.5
3.85
4.43
3.97
0.89
5.41
4.49
4.88
2.21
1.97
6.82
9.8
5.96
7.65

1. Define the population parameter in one sentence.
2. State the null and alternative hypotheses using correct notation.
3. State the significance level for this problem.
4. ) Calculate the test statistic “by-hand.” Show the work necessary to obtain the value by typing your work and provide the resulting test statistic. Do not round during the calculation. Then, round the test statistic to two decimal places after you complete the calculation.
5. State the p-value using the output provided in part (d). Use four decimal places for the p-value. In addition, state the degrees of freedom.
6. State whether you reject or do not reject the null hypothesis and the reason for your decision in one sentence.
7. State your conclusion in the context of the problem (i.e. interpret your results and/or answer the question being posed) in one or two complete sentences.
8. Construct a 99% confidence interval.
9. Explain the connection between the confidence interval and the hypothesis test in this problem (discuss this in relation to the decision made from your hypothesis test and connect it to the confidence interval you constructed in part (k)). Answer this question in one to two sentences.

Answer 1

a) Let $\mu$ denotes the mean file size of Data Analysis 2.

To test null hypothesis Ho :u =5.00 against alternative hypothesis H1 : # 5.00 Let ã denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, ã = 5.3895, $a = 5.6761, s = 2.3825, n = 22 The test statistic can be written as : t- vn (m – 5.00) which under Ho follows a t distribution with n - 1 df. We reject H, at 0.010 level of significance if P-value < 0.010 or if tobs | > to.005,n–1 Now, The value of the test statistic = 0.766912 associated degrees of freedom = 21 The critical value =+ 2.831360 P – value = P(|tn-11 > tobs) = 2 * P(t21 > 0.766912) = 2 *(1 – 0.774165) = 0.451670 Since p-value > 0.01 and tobs | * tcritical = 2.831360, so We fail to reject Ho at 0.01 level of significance Hence, we can conclude that the population mean is not significantly different from 5.00.

Conclusion : There is not sufficient evidence to conclude that the mean file size of Data Analysis 2 is different from 5 megabytes.

Here, ī = 5.390, s? = 5.6761, s = 2.3825, n = 22 The margin of error is: S ME= ME = r * n = 2,382 + 60,2 2,982 * 2.831360 = 1.438161 – 1.438 - * t 1-0.990 0.990 21 = 2.382 R 2.382 — ** 2.831360 = 1.438161 ~ 1.438 * 10.005,21 = 2 V22 122 a 99 percent confidence interval population mean u is = *t 1-0.990 21 = 2 ,4 vn 2.382 = 5.390 + * t0.005,21 V22 = 5.390 + 4 *2.831360 0.390+ 2.382 V22 = 5.390 – 1.438161 = (3.951384, 6.827707) → (3.95, 6.83)

Since the confidence interval includes 5, so at 95% level of confidence or at 5% level of significance, we can conclude that the mean file size of Data Analysis 2 is not significantly different from 5 megabytes.