Save Homework: Homework Chapter9 Score: 0.57 of 4 pts 7017(6 complete) HY Score:

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Save Homework: Homework Chapter9 Score: 0.57 of 4 pts 7017(6 complete) HY Score: 34.51 %, 8.97 of 26 pts 9459-T Question Help 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 t were made available at a substantially reduced cost. Data are collected from a random sample of 600 subscribers. The results indicate that 134 of the subscribers below a.Atee 005 level of signfcance,is there evidence that Determine the null hypothesis, Ho, and the altemative hypothesis, H1 Id upgrade anew cellphone at an od cost Red ngte pri will be profitable if at last 20% of the subscribe swold prade. Complete parts a and b more than 20% of the astomers would upgrade to a new cellphone at a reduced cost? A. Ho: 0.20 > 0.20 OD, Ho: 0.20 H1: #0.20 OC. Hoi 20.20 H1: 020 What is the test statistic? zsm"(Type an integer or a decimal. Round to two decimal places as needed.)

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P > 0.20

Alternative hypothesis: P < 0.20

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.01633

z = (p - P) /

z = 1.43

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than 1.43. We use the Normal Distribution Calculator to find P(z < 1.43).

Thus, the P-value = 0.9236

Interpret results. Since the P-value (0.9236) is greater than the significance level (0.05), we have to accept the null hypothesis.

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