Mathematics, 20.12.2019 03:31 honeylebling

Suppose the mean wait-time for a telephone reservation agent at a large airline is 40 seconds. a manager with the airline is concerned that business may be lost due to customers having to wait too long for an agent. to address this concern the manager develops new airline reservation policies that are intended to reduce the amount of time an agent needs to spend with each customer. a random sample of 250 customers results in a sample mean of 39.3 seconds with a standard deviation of 42 seconds using alpha = 0.05 level of significance, do you believe the new policies were effective in reducing wait time? do you think the results have any practical significance?
determine the null and alternative hypotheses h_0: 40 seconds h_1: 40seconds calculate the test statistic t_0 = calculate the p- value. p-value = state the conclusion for the test. a. reject h_0 because the p-value is greater than the alpha = 0.05 level of significance. b. reject h_0 because the p-values is less than the alpha = 0.05 level of significance. c. do not h_0 because the p-value is less than the alpha = 0.05 level of significance. d. do not reject h_0 because the p-value is greater than the alpha = 0.05 level of significance. state the conclusion in context of the problem. there sufficient evidence at the alpha = 0.05 level of significance to conclude that the new politics were effective.

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Answer from: jaylen2559

Null hypothesis: $\mu \geq 40$

Alternative hypothesis : $\mu$

$t=\frac{39.3-40}{\frac{42}{\sqrt{250}}}=-0.264$

$p_v =P(t_{(249)}$

D. Do not reject H_0 because the P-value is greater than the alpha = 0.05 level of significance.

We dont't have enough evidence to conclude that the time is reducing with the new policies since we fail to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=39.3$ represent the mean wait-time for a telephone reservation agent at a large airline

s=42 represent the sample standard deviation

n=250 sample size

$\mu_o =40$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

Is a one tailed left test.

What are H0 and Ha for this study?

Null hypothesis: $\mu \geq 40$

Alternative hypothesis : $\mu$

Compute the test statistic

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{39.3-40}{\frac{42}{\sqrt{250}}}=-0.264$

Calculate the p value

First we need to fidn the degrees of freedom given by df=n-1=250-1=249

Since is a one side left tailed test the p value would be:

$p_v =P(t_{(249)}$

D. Do not reject H_0 because the P-value is greater than the alpha = 0.05 level of significance.

State the conclusion in context of the problem. There sufficient evidence at the alpha = 0.05 level of significance to conclude that the new politics were effective.

We dont't have enough evidence to conclude that the time is reducing with the new policies since we fail to reject the null hypothesis at 5% of significance.

Answer from: Quest

2500

step-by-step explanation:

Answer from: Quest

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!

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Suppose the mean wait-time for a telephone reservation agent at a large airline is 40 seconds. a man...