Question

Asked Mar 18, 2019

147 views

Step 1

**Find the distribution that has to be used to construct the confidence interval in the given situation:**

It is given that, a sample of size 60 with mean 60.1 and standard deviation 6.7 is used to estimate the population mean.

The sample size is *n* = 60,

The sample mean is *x*–bar = 60.1,

The sample standard deviation is *s * = 6.7.

The required conditions for using *t *distribution to estimate the confidence interval about mean is given below:

- The population standard deviation σ must be unknown.
- Either the sample size (
*n*) must be greater than 30 or the population must be normally distributed.

Here, the population standard deviation is unknown. The data just tells about the sample standard deviation.

Moreover the sample size is *n *= 60 (Which is greater than 30).

Hence, the two conditions are satisfied.

Thus, the *t*–distribution is used to construct the confidence interval for the population mean in the given situation.

The approximate 100*(1–α)% confidence interval for the population mean will be obtained using the formula given below:

Step 2

**Obtain the critical value:**

The required confidence level is 100*(1–α)% = 99%.

From the given common *t-*values using *t-*distribution the critical value is obtained as 2.662 from the calculation given below:

Step 3

**Find the 95% confidence interval for the population mean:**

Here, *n* = 60, *x*–bar = 60.1, *s* = 6.7.

The critical value is *t*(α/2, *n–*1) = *t*0.005,59 = 2.662.

Denote the population mean as µ.

The 95%...

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