Question

In one year, an online travel agency reported that summer travelers booked their airline reservations an average of 72.4 days in advance. A random sample of 40 summer travelers in a later year was selected, and the number of days travelers booked their airline reservations in advance was recorded. The data are shown in the accompanying table. Complete parts a through c below. Click the icon to view the data table. a. Perform a hypothesis test using a = 0.01 to determine if the average number of days reservations are booked in advance has changed between the years. Determine the null and alternative hypotheses. Choose the correct answer below. OA. Hp: #72.4 and Hy: p=72.4 B. Ho: x=72.4 and Hy: >72.4 OC. Ho: 272.4 and H <72,4 OD. How=72.4 and Hu*72.4 Determine the test statistic test statistic- (Round to two decimal places as needed.) Determine the critical value(s) The critical value(s) isare) (Round to three decimal places as needed. Use a comma to separate answers as needed.)

Answer 1

a)

Null Hypothesis(H₀): $\mu$ = 72.4

Alternative Hypothesis(H₁): $\mu$ $\neq$ 72.4.

Correct Answer is (D)

Sample size(n) = 40

Suppose that airline reservation of x days in advance is normnal random variable with unknown mean $\mu$ and unknown variance $\sigma$ ² .We use to test the hypothesis that the mean equals a standard value 72.4, that is

H₀: $\mu$ = 72.4

H₁: $\mu$ $\neq$ 72.4.

where level of significance is 0.01

the test statistics is

t₀ = ($\bar{}$$\bar{x}$ - 72.4)$\div$ (s$\div$$\sqrt{n}$)

s = sample standard deviation = 9.54

$\bar{x}$ = sample mean = 74.15

the test statistics is the t distribution with 40-1 =39 degree of freedom.

t₀ = 1.16016 ~ 1.16

critical value t _0.005,40= 2.704

- 2.704 < t₀< 2.704

(D) Do not reject the null hypothesis. The data do not provide sufficient evidence to conclude that the average number of days has changed between the years.  

b)

Use minitab

One-Sample T: x

Descriptive Statistics

N	Mean	StDev	SE Mean	99% CI for μ
40	74.15	9.54	1.51	(70.06, 78.24)

μ: mean of x

Test

Null hypothesis	H₀: μ = 72.4
Alternative hypothesis	H₁: μ ≠ 72.4

T-Value	P-Value
1.16	0.253

P- value = 0.253

c)

The P-value is the probability that the test statistic will take on a value that is at least as extreme as the observed value of the statistic when the null hypothesis H₀ is true.

correct answer is (C).