Question

Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic, P-value, and critical values. Express to a maximum of 3 decimal places.

Claim: The mean time between uses of a TV remote control by males during commercials equals 5.00 seconds.

Sample data: n = 81, x = 5.25 sec, s = 2.90 sec. The significance level is α = 0.01.

(a) test statistic
(b) P-value
(c) critical values:	(lower value),  (higher value)

Answer the following questions based upon the previous exercise above.

(a) What was the original claim?

μ ≠ 5.00μ < 5.00    μ = 5.00μ > 5.00μ ≤ 5.00μ ≥ 5.00

(b) What was the null hypothesis?

H₀: μ ≥ 5.00H₀: μ ≤ 5.00    H₀: μ ≠ 5.00H₀: μ > 5.00H₀: μ < 5.00H₀: μ = 5.00

(c) What was the alternative hypothesis?

H₁: μ < 5.00H₁: μ > 5.00    H₁: μ ≠ 5.00H₁: μ ≥ 5.00H₁: μ = 5.00H₁: μ ≤ 5.00

(d) What was the final conclusion?

Reject the null hypothesis. Sample provides enough evidence to reject the claim.Reject the null hypothesis. Sample does not provide enough evidence to reject the claim.    Do not reject the null hypothesis. Sample provides enough evidence to reject the claim.Do not reject the null hypothesis. Sample does not provide enough evidence to reject the claim.

Answer 1

Given

X_bar = 5.25

s = 2.90

n= 81

Claim : the mean time between uses of TV remote control by males during commercial equal 5.00 sec.

1) null hypothesis:

H0 : = 5.00

2) alternative hypothesis:

H1 : 5.00

3) test statistic Z

Z = (x_bar - ​​​​​​) /(s/sqrt(n))

= (5.25-5)/(2.90/sqrt(81))

= 0.78

Z = 0.78

4) p value for Z test statistic is 0.4354

P value = 0.4354

5) conclusion:

P value (0.4354) is greater than 0.01 level of significance that fail to reject null hypothesis. Sample does not provide enough evidence to reject the claim.