A report in Ohio states that the mean cost for eye insurance is \$150 a year with a standard deviation of \$26. (population standard deviation). Test if the mean cost for eye insurance in KY is different that OH. A random sample of 80 people is taken with mean cost found to be \$ 180. Does this differ from what was reported for OH? a - .05.  Create an Ho and Ha. Select appropriate test. Decision Rule. Calculate the test statistics. Conclusion with comparison to alpha or critical value.this would be a two tailed testt test

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Asked Mar 11, 2019
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A report in Ohio states that the mean cost for eye insurance is \$150 a year with a standard deviation of \$26. (population standard deviation). Test if the mean cost for eye insurance in KY is different that OH. A random sample of 80 people is taken with mean cost found to be \$ 180. Does this differ from what was reported for OH? a - .05.  Create an Ho and Ha. Select appropriate test. Decision Rule. Calculate the test statistics. Conclusion with comparison to alpha or critical value.

this would be a two tailed test

t test

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Expert Answer

Step 1

Test whether the population mean cost of eye insurance is equal to 150 or not:

Denote the population mean as μ, and the population standard deviation as σ.

Given that, the population mean cost of eye insurance is μ = 150, the population standard deviation in cost of eye insurance is σ = 26.

The hypotheses are given below:

Null hypothesis:

H0 : μ = 150

That is, the population mean cost of eye insurance is equal to 150.

Alternative hypothesis:

H1 : μ ≠ 150 (Two tail test).

That is, the population mean cost of eye insurance is not equal to 150.

Step 2

Obtain the test statistic value.

The sample mean cost of eye insurance is x-bar = 180. The sample size is n = 80.

The test statistic value is obtained as 10.32 from the calculation given below:

Step 3

Obtain the critical value:

The level of significance is α = 0.05.

From the standard normal distribution a...

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