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What is the test statistic? z=____ (Round to two decimal places as needed.) What

ID: 3326708 • Letter: W

Question

What is the test statistic?

z=____

(Round to two decimal places as needed.)

What is the P-value?

P-value=_____

(Round to four decimal places as needed.)

Score: 0.2 of 1 pt 18 of 35 (23 complete) Hw Score: 60.43%, 21.15 of 35 pts &8.3.34-T Question Help Consider a flight to be on time if it arrives no later than 15 minutes after the scheduled arrival time. Negative arrival times correspond to flights arriving earlier than their scheduled arrival time. Use the sample data to test the daim that 79.7% of flights are on time. Use a 0.025 significance level and the P-value method to answer the following questions. Click on the icon to view the table of arrival delay times. What are the null and alternative hypotheses? O A. Ho:p- 0.797 O C. Ho:P> 0.797 OE. Ho:p#0.797 H1:p> 0.797 H1 : p = 0.797 H1 : p = 0.797 O B. Ho:p=0.797 H1 p

Explanation / Answer

Solution:-

x = 35

n = 40

p = 35/40

p = 0.875

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

Null hypothesis: P = 0.797
Alternative hypothesis: P 0.797

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.025. 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 ]

= 0.0636
z = (p - P) /

z = 1.23

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 two-tailed test, the P-value is the probability that the z-score is less than - 1.23 or greater than 1.23.

Thus, the P-value = 0.2186

Interpret results. Since the P-value (0.2186) is greater than the significance level (0.025), we cannot reject the null hypothesis.

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