Question

Assume a media agency reports that it takes television streaming service subscribers in the United States an average of 6.03 days to watch the entire first season of a television series, with a standard deviation of 3.98 days. Suppose Elizabeth, an analyst for an online television and movie streaming service company, wants to determine if her company's customers exhibit similar viewing rates for their series offerings.

She formulated the null hypothesis ?0:?=6.03 daysH0:μ=6.03 days and the alternative ?1:?≠6.03 daysH1:μ≠6.03 days, where ?μ is the mean number of days that it takes for customers to watch the first season of a series. She conducted a one-sample ?-z-test based on data collected from a simple random sample of 820 customers who watched the first season of a television series from the company database of over 20,000 customers that qualified. She assumed that the population standard deviation is ?=3.98 days.

She compiled the summary statistics shown in the table.

Sample size	Sample mean	Standard error
?n	?⎯⎯⎯x¯	SE
820	6.33	0.14

Compute the ?-value for Elizabeth's hypothesis test directly using a normal CDF function on a TI calculator or software. You may find some software manuals helpful. Provide your answer with precision to four decimal places. Avoid rounding until the final step.

?-value =

Answer 1

The hypotheses are Ho : u=6.03 vs HQ:#6.03. The test is two-sided. The sample size is n=820. The sample mean is 7 = 6.33. The standard error is SE = 0.14. The test statistic is calculated as follows: I - Ho SE 6.33-6 0.14 z= 2.357143 The P-value is calculated as follows: P-value = P(Z > 2.357143) + P(Z<-2.357143) = 2P(Z<-2.357143) (Since P(Z > 2.357143)=P(Z<-2.357143)) = 2(0.009208077) (using pnorm(-2.357143) function in R) =0.01841615 P-value = 0.0184

Answer 2

Using Ti - 84 calculator,

Output :

Z-Test(6.03, 3.98, 6.33, 820, 0) 2->2.15846548693 p->0.030891658

From provided information

n = 820

$\bar{x} = 6.33$

$\sigma= 3.98$

$SE= \frac{\sigma}{\sqrt{n}}$

SE = 0.14

We want to test the following null and alternative hypothesis

Ho:p = 6.03

$H1 :\mu \neq6.03$

Test statistic ( Z ) = $\frac{\bar{x}- \mu}{SE}$

Z = ( 6.33 - 6.03)/0.14

Z = 2.1584

P- value = P( | Z | > 2.1584)

​​​​​​P- value = 0.0309

Thank You!

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