Assume a media agency reports that it takes television streaming service subscribers in the U.S. an average of 6.33 days to watch the entire first season of a television series, with a standard deviation of 3.98 days. Scarlett is an analyst for an online television and movie streaming service company that targets to the 18-50 age bracket. She wants to determine if her company's customers exhibit shorter viewing rates for their series offerings. Scarlett plans to conduct a one-sample z-test with significance level of α = 0.05, to test the null hypothesis, H0: μ =633 days, against the alternative hypothesis, HI : μ < 6.33 days. The variable μ is the mean amount of time, in days, that it takes for customers to watch the first season of a television series.
Scarlett selects a simple random sample of 840 customers who watched the first season of a television series from the company database of over 25,000 customers that qualified. She compiles the summary statistics shown in the table. Sample size Sample mean Standard error
n x SE
840 6.14 0.1373
Required:
1. Compute the p-value for Scarlett's hypothesis test directly using a normalcdf function on a TI calculator or by using software.

Respuesta :

Answer:

The p-value is 0.0832.

Step-by-step explanation:

Given:

The hypothesis is:

H0: μ =633 vs. HI : μ < 6.33

The significance level is a = 0.05.

The sample size is, n=840, sample mean, x = 6.14 and standard error, SE = 0.1373.

Compute the test statistic:

X-u/S.E.

(6.14-6.33)/0.1373

=-1.38383

-1.384

Use the excel function" =NORMSDIST(-1.384)" to compute the p-value see the attached file.

Therefore,

The p-value is 0.0832.

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