A certain vehicle emission inspection station advertises that the wait time for

Question

A certain vehicle emission inspection station advertises that the wait time for customers is less than

7

minutes. A local resident is skeptical and collects a random sample of

49

wait times for customers at the testing station.

SheShe

finds that the sample mean is

5.53

minutes, with a standard deviation of

2.5

minutes. Does the sample evidence support the resident's skepticism? Use the

alphaequals=0.05

level of significance to test the advertised claim.

What are the correct hypotheses for this test?

A.

H0:

muequals=7

minutes

H1:

munot equals7

minutes

B.

H0:

muequals=7

minutes

H1:

muless than<7

minutes

C.

H0:

muequals=7

minutes

H1:

mugreater than>7

minutes

D.

H0:

muless than<7

minutes

H1:

muequals=77

minutes

E.

H0:

mugreater than>7

minutes

H1:

muequals=7

minutes

F.

H0:

munot equals7

minutes

H1:

muequals=7

minutes

What is the value of the test statistic?

t 0t0

z 0z0

equals=nothing (Round to two decimal places as needed.)

What is the P-value?

P-valueequals=nothing

(Round to three decimal places as needed.)Use the

alphaequals=0.05

level of significance to test the advertised claim.Since the P-value is

greater

less

than

alpha,

reject

do not reject

the null hypothesis. There

is

is not

sufficient evidence to conclude that the mean wait time is less than

77

minutes. In other words, the evidence

does not support

supports

the resident's skepticism.

Explanation / Answer

Solution :-

Given, Samples = n = 49

Sample mean = 5.53

Standard deviation = 2.5

Level of significance = = 0.05

Also, as it says certain vehicle emission inspection station advertises that the wait time for customers is less than 7 minutes. For hypothesis we choose,

B. Null hypothesis, H0: mu equals = 7 minutes

Alternate hypothesis, H1: mu less than < 7 minutes

Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n) = 2.5 / sqrt(49) = 0.357
DF = n - 1 = 49 - 1 = 48
t = (x - ) / SE = (5.53 - 7)/0.357 = - 4.118

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Here the sample size is more than 30 so we consider the z test given as,

Z = (X - ) /
Z = (5.3 - 7) / 2.5

Z = - 0.68

The corresponding P-Value is 0.248252.

Since the P-value (0.25) is greater than the significance level (0.01), we cannot reject the null hypothesis.

There is not sufficient evidence to conclude that the mean wait time is less than 7 minutes. In other words, the evidence does not support the resident's skepticism.

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