Answer to Question #334003 in Statistics and Probability for Hansi

Question

A research firm conducted a study in 2014 and found that 40% of Internet users

received more than 10 emails per day. A similar study on the use of email

repeated in 2016.

i. Formulate the hypotheses that can be used to determine whether the

proportion of Internet users receiving more than 10 emails per day

increased in 2016.

i i. In a sample of 450 Internet users found 190 receiving more than 10 emails per

day, what is P-value?

iii. At a = 0.05, what is your conclusion?

(C) Consider the following hypothesis test:

Ho; = 10

H A : not equal to10

A sample of 50 provided a sample mean of 9.80. The population standard deviation is

4.

i. Compute the value of the test statistic.

i i. What is P-value?

iii. At a = 0.10, what is your conclusion?

Answer 1

(A)

i. The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p\le 0.4$

$H_a:p>0.4$

This corresponds to a right-tailed test, for which a z-test for one population proportion will be used.

ii.

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{\dfrac{190}{450}-0.4}{\sqrt{\dfrac{0.4(1-0.4)}{450}}}\approx0.96225$$

The p-value is $p =P(Z>0.96225)= 0.167962$

iii.Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a right-tailed test is $z_c = 1.6449.$

The rejection region for this right-tailed test is $R = \{z: z > 1.6449\}.$

Since it is observed that $z = 0.96225 \le1.6449= z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p = 0.167962,$ and since $p = 0.167962 \ge 0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is greater than 0.4, at the $\alpha = 0.05$ significance level.

(C)

The following null and alternative hypotheses need to be tested:

$H_0:\mu=10$

$H_a:\mu\not=10$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

i. The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{9.8-10}{4/\sqrt{50}}=-0.353553$$

ii. The p-value is $p =2P(Z<-0.353553)= 0.723674.$

iii.Based on the information provided, the significance level is $\alpha = 0.10,$ and the critical value for a two-tailed test is $z_c = 1.6449.$

The rejection region for this two-tailed test is $R = \{z: |z| > 1.6449\}.$

Since it is observed that $|z| =0.353553 \le1.6449= z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p = 0.723674,$ and since $p = 0.723674 \ge 0.10=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 10, at the $\alpha = 0.10$ significance level.