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StatisticsQ&A LibraryA 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 testQuestion

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

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:*

*H*_{0} : μ = 150

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

*Alternative hypothesis:*

*H*_{1} : 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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