Statistic Problems

9.1 Because Z stat= -0.76<-1.96, do not reject H0 because the test statistic is not less than the critical value.

9.2 Because Z stat= 2.21> 1.96, reject because the test statistic is more than the critical value.

9.28

A. -2.1448< tstat= 1.6344< 2.1448 do not reject H0. Not enough evidence that show that the amount spent is different from $6.50.

B. P-Value= 0.1245. This mean the probability of rejecting the null hypothesis (H0) of the sample.

C. The assumption that I would make is that its normally distributed.

D. The distribution is symmetric because both the mean and the median value are close, due to a small sample size. The boxplot are slightly skewed so normality assumptions are not seriously violated.

9.29

A. Ho = 45 hours

Ha ≠ 45 hours

α= 0.05

Test Statistic: = =

Decision Rule: Reject H0 if tstat is less than – 2.056 or greater than or 2.056

Decision: Do not reject H0. (t statistic -0.2283 is less than t critical 1.729)

At the 5 percent significance level, the data do not provide sufficient evidence to conclude that the amount of mean processing insurance application time differed from that in the past.

B. We assume that the population is approximately normally distributed and a random sample.

D. The assumption would not be valid because the sample does not follow normal distribution.

9.35

A. Step 1 H0 Sample mean = 144 min

H1 Sample mean is not equal to 144 min

Step 2 0.05

Step 3 1 sample t test

Test statistic 1.224674365

xbar 175.2666667

s 139.8368343

n 30

mu 144

Step 4 0.230553269 p_value

Step 5 p_value > 0.05 We fail to reject the H0

Therefor there is no evidence that the population mean time spent on the internet is different from 144 minutes.

B. The t-test assumes that the data would represent a random sample form the population that is normally distributed.

C. The t-test is used when the population standard deviation is unknown, you can use the t-test to assume the data represents a random sample form a population is normally distributed, and also the t-test will make a good assumptions for sampling distribution of the mean when the standard deviation is unknown.

D. All the assumptions is B and C are valid.

9.74

A. Tstat= -1.69> -1.7613, do not reject the H0

B. Normally distributed

D. Pretty much normally distributed.

E. There is no evidence that shows that the waiting time is less than 5 minutes.

9.75

A. Tstat= 1,689.96<1,756.84, Reject the H0

B. That is normally distributed

D. Yes the assumptions are needed because the value as close to each other.

9.76

A. Tstat= -1.47>-1.6896, do not reject the H0

B. p-value= 0.0748. That means the probability of obtaining the tstat ,-1.47, or more is extreme 0.0748 if the null hypothesis is true.

C. Tstat- -3.10<-1.6973, reject H0

D. p-value= 0.0021. That mean if the null hypothesis is true, the probability of getting a tstat of -3.10 or more extreme is 0.0021.

E. Normally distributed

G. The assumptions in order to conduct the t tests in a and c is valid. They have very large sample sizes and the results from the t test are insensitive and unnormal.

9.77

A. 1.4*10^-37<0.05, Null hypothesis should be rejected H0

B. P-value indicates that the null hypothesis is true and a mean that differed by 26 pounds would be expected of 1.4*10^37 samples

C. There is no evidence that show that the population mean weight is different from 3700.

D.P-value= 0.12. If the null hypothesis is true, mean by 4 or more would be expected for 12% of the samples.

E. Due to the large sample sizes the populations can be normally distributed so the normality assumption is not a concern.

9.78

A. Tstat= -3.2912, reject H0

B. P-value= 0.0012. That means the probability of getting a tstat above +3.2912 or below the -3.2912 is 0.0012

C. Tstat= -7.9075, reject H0

D.p-value= 0.00. That means the probability of getting a tstat above +7.9075 or below-7.9075 is 0.00.

E. Due to the large sample sizes the populations can be normally distributed so the normality assumption is not a concern.