Mathematics, 22.02.2020 04:36 solo4360

A call center company requires it's operators to answer the phone in an average time of 0.17 minutes or less. A sample of 32 actual operator times were collected and the results were a mean time of 0.173 minutes and a standard deviation of 0.0183 min. Is this sample mean response time statistically higher than the standard? What can you conclude? Assume alpha = 0.05

Select one:

a. There is statistical evidence that the mean response time exceeds the standard.

b. There is no statistical evidence that the mean response time exceeds the standard.

c. The mean response time is exactly equivalent to the standard.

d. The mean response time exceeds the standard by alpha.

Answers: 1

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Answer from: nyha90

$t=\frac{0.173-0.17}{\frac{0.0183}{\sqrt{32}}}=0.927$

df=n-1=32-1=31

$p_v =P(t_{31}0.921)=0.182$

If we compare the p value and a significance level given $\alpha=0.05$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis. And the best conclusion would be:

b. There is no statistical evidence that the mean response time exceeds the standard.

Step-by-step explanation:

Data given and notation

$\bar X=0.173$ represent the average score for the sampe

s=0.0183 represent the sample standard deviation

n=32 sample size

$\mu_o =0.17$ 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.

We need to apply a one lower tailed test.

What are H0 and Ha for this study?

Null hypothesis: $\mu \leq 0.17$

Alternative hypothesis : $\mu 0.17$

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:

$t=\frac{0.173-0.17}{\frac{0.0183}{\sqrt{32}}}=0.927$

Give the appropriate conclusion for the test

First we need to find the degrees of freedom given by:

df=n-1=32-1=31

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

$p_v =P(t_{31}0.921)=0.182$

Conclusion

b. There is no statistical evidence that the mean response time exceeds the standard.

Answer from: Quest

17 is c. congruent means equal in size

Answer from: Quest

the correct answer is you spend $55.78 more in a brand store that buy ing in a normal store.

step-by-step explanation:

brand = spend $ 135.85 total

store = spend $80.07 total

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