Question

Your instructor (Katie) believes that the average time of taking this exam is equal to 140 minutes. A sample of 4 students was taken and the following exam times were obtained. Assume the distribution of the population is normally distributed. Sample: 150, 150, 180, 170 (Sample Std. Dev = 15) At a 0.1 level of significance, it can be concluded that the mean of the population is: O a significantly less than 140 O b. not significantly different than 140 O c. significantly greater than 140 O d. significantly different than 140

Answer 1

Sample size, n= 4

Sample mean,
T
= $\frac{150+150+180+170}{4}$ = $\frac{650}{4}$ = 162.5

Sample standard deviation, s= 15 (given)\

Null hypothesis, H0 : $\mu$ = 140 min

Alternate hyothesis, Ha: $\mu \neq$ 140 min

Test statistic, t= $\frac{\bar{x}-\mu}{s/\sqrt{n}}$ = $\frac{162.5-140}{15/\sqrt{4}}$ = 2.9867

critical value, t_{ 3, 0.1/2} = $\pm$ 2.3534

Now, since the test statistic is greater than critical value, That means that the test statistic lie in the critical region.

Hence we will reject the null hypothesis. Hence the mean of the population is significantly different than 140.

Answer- d. significantly different than 140.