Question

Assume a media agency reports that it takes television streaming service subscribers in the U.S. an average of 6.14 days to watch the entire first season of a television series, with a standard deviation of 4.03 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 ?-test, with significance level of ?=0.05, to test the null hypothesis, ?0:?=6.14 days, against the alternative hypothesis, ?1:?<6.14 days. The variable ? 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 825 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 ? ?⎯⎯⎯ SE 825 5.84 0.1403 Compute the ?-value for Scarlett's hypothesis test directly using a normalcdf function on a TI calculator or by using software. You may find some software manuals helpful. Provide your answer with precision to four decimal places. Avoid rounding until the final step. Select the accurate statement regarding Scarlett's test decision if she tests at a significance level of ?=0.05. Scarlett should fail to reject the null hypothesis because the ?-value is greater than the value, ?=0.05. Scarlett should fail to reject the null hypothesis because the ?- value is less than the value, ?=0.05 . Scarlett should reject the null hypothesis because the ?-value is less than the value, ?=0.05. Scarlett should reject the null hypothesis because the ?-value is greater than the value, ?=0.05. Scarlett should reject the null hypothesis because the ?-value is less than the value, ?2=0.025.

Answer 1

It is given that

sample mean (x) = 5.84

mean (population) 6.14

standard error = 0.1403

sample size is n = 825

using normalcdf(lower bound, upper bound, mean, standard error)

where lower bound = -E99, upper bound = 5.84, mean= 6.14 and standard error = 0.1403

setting the values

= normalcdf(-E99,5.84,6.14,0.1403)

= 0.0162

it is clear that p value is less than 0.05 level of significance.

Therefore, rejecting the null hypothesis

Correct answer is "Scarlett should reject the null hypothesis because the ?-value is less than the value, ?=0.05"