Question

A sample of 400 items is taken from a population whose standard deviation is 10. The mean of sample is 40. Test whether the sample has come from a population with mean 38. Also calculate 95% confidence interval for the population.

Answer 1

Step-by-step explanation:

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to your Z score and level of significance.

95% = mean ± 1.96 SEm

Answer 2

Answer:

95% confidence interval  = 39.02, 40.98

Step-by-step explanation:

Given:- n = 400, x = 40, µ = 38, σ = 10.

To Find:- Test if the sample has come from a population with mean 38 and also calculate 95% confidence interval for the population.

Solution:-

We know that,

         The test statistic, Z = (x - µ) / (σ/√n)

          $Z_{\alpha }$ = 1.96

Putting all the given values in the above mentioned formula, we get

Z = (40 - 38) / (10/√400)

  = 2 / (10/20)

  = 2 × 2

  = 4.

∵ Z = 4 > 1.96 ($Z_{\alpha }$)

∴ We reject the Null hypothesis i.e.  µ = 38.

Therefore, the sample is not from a population with mean 38.

95% confidence interval  = (x ± 1.96( σ/√n ))

                                          = (40 ± 1.96( 10/√400 ))

                                          = 40 ± 1.96 × 0.5

                                          = 40 ± 0.98

                                          = 39.02, 40.98

Therefore, 95% confidence interval for the population is 39.02 and 40.98.

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