Question

A cellphone provider has the business objective of wanting to determine the proportion of subscribers who would upgrade to a new cellphone with improved features if it were made available at a substantially reduced cost. Data are collected from a random sample of 500 subscribers. The results indicate that 110 of the subscribers would upgrade to a new cellphone at a reduced cost. Reducing the price will be profitable if at least 20% of the subscribers would upgrade. Complete parts (a) and (b) below. a. At the 0.05 level of significance, is there evidence that more than 20% of the customers would upgrade to a new cellphone at a reduced cost? Determine the null hypothesis, Ho, and the alternative hypothesis, H. O A. Ho XS 0.20 H:>0.20 OB. Ho: 20.20 H:<0.20 OD. H : **0.20 H: = 0.20 O C . H: =0.20 Hi**0.20 What is the test statistic? ZSTAT = (Type an integer or a decimal. Round to two decimal places as needed.) What is the p-value? The p-value is ! (Type an integer or a decimal. Round to three decimal places as needed.) What is the final conclusion? the null hypothesis. There sufficient evidence that the percentage of people who would upgrade to a new cellphone at a reduced cost is 20%. b. How would the manager in charge of promotional programs concerning cellphone upgrades use the results in (a)? The manager suggest reducing the price.

Answer 1

The following information is provided: The sample size is N = 500 , the number of favorable cases is X = 110 , and the sample proportion is $\bar p = \frac{X}{N} = \frac{ 110}{ 500} = 0.22$ , and the significance level is α=0.05

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: p ≤ 0.2

Ha: p > 0.2

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

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is z_c = 1.645

(3) Test Statistics

The z-statistic is computed as follows:

$z = \frac{\bar p - p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{ 0.22 - 0.2 }{\sqrt{ 0.2(1- 0.2)/500}} = 1.118$

(4) Decision about the null hypothesis

Since it is observed that z = 1.118 < z_c = 1.64 , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.1318 , and since p = 0.1318 ≥ 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is greater than p0​, at the α=0.05 significance level.

Do not reject the null hypothesis. There there is not sufficient enough evidence that the percentage of people who would upgrade to a new cell phone at a reduced cost is more than 20%

The manager should not suggest reducing the price