The data collected for the 30 students is presented below in the order it was re

Question

The data collected for the 30 students is presented below in the order it was received: 2, 1, 0, 4, 2, 3, 3, 0, 0, 0, 2, 3, 2, 1, 1, 5, 0, 4, 1, 0, 0, 2, 5, 1, 3, 2, 0, 2, 0, 3

The sample mean for times physical activity is 1.73

The median for times physical activity per week is 2

The mode for times of physical activity is 0

The highest amount of weekly physical activity reported for that week was 5 and the lowest was 0. Therefore, the range is 5.

The sample variance for times of physical activity is 2.49 times^2.

The sample standard deviation for times of physical activity is 1.55.

The coefficient of variation for times of physical activity is 89.55%.

Clearly complete one-sample t test of hypothesis when the population standard deviation is unknown using a population mean of 3. Use either the "6 Steps in Hypothesis Testing" using critical values or "The 5 Step p-value approach to Hypothesis Testing".

EXAMPLE of format:

Step 1: Ho: = 70 ; H1: 70

Step 2: LOS = 0.05

Step 3: use t statistic since not known

Step 4: reject Ho if tSTAT< -2.045 or tSTAT> +2.045 (found using T.INV.2T)

Step 5: tSTAT= (75.77 –70) / (10.95 / 30 ) = 5.77 / 2 = 2.885

Step 6: reject Ho since tSTAT> 2.045; also, p-value = 0.003656 < of 0.05.

Conclusion: There is sufficient statistical evidence to conclude that the average temperature of all U.S. cities was not 70F on October 5, 2015.

Explanation / Answer

Solution:

The data collected for the 30 students is presented below in the order it was received: 2, 1, 0, 4, 2, 3, 3, 0, 0, 0, 2, 3, 2, 1, 1, 5, 0, 4, 1, 0, 0, 2, 5, 1, 3, 2, 0, 2, 0, 3.

The sample mean for times physical activity is 1.73 , ie X_bar = 1.73

The sample standard deviation for times of physical activity is 1.55 , ie s = 1.55

Step 1: H0: = 3 ; H1: 3

Step 2: LOS = 5% = 0.05

Step 3: Here we use t statistic, since not known

Step 4: We reject H0 iff tSTAT< -2.045 or tSTAT> +2.045 (found using T.INV.2T) (t0.025,(n-1) = t0.025,29 = 2.045)

Step 5: tSTAT= ((1.73-3) / (1.55 /30)) = - 4.4878

Step 6: We reject H0, since tSTAT = -4.4878 < -2.045; also, p-value = 0.000105016 < of 0.05.

Conclusion: Since tSTAT = -4.4878 < -2.045; also, p-value = 0.000105016 < of 0.05, we thus reject H0 at 5% level of significance and conclude on the basis of the given data that the population mean for times physical activity is not 3 (ie, it is statistically different from 3).

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.