Question

A random sample of 9 values from normal population gave the values :66,68,59,67,64,66,63,62,61. Discuss the suggestion that the mean in the population is 60, given σ=3. Use α=0.05
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Answer 1

We conclude that the mean in the population is not equal to 60.

Step-by-step explanation:

We are given a random sample of 9 values from the normal population gave the values: 66, 68, 59, 67, 64, 66, 63, 62, 61.

Also, σ = 3 and the level of significance $\alpha$ = 0.05.

Let $\mu$ = population mean

So, Null hypothesis, $H_0$ : $\mu$ = 60      {means that the mean in the population is 60}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 60     {means that the mean in the population is not equal to 60}

The test statistics that would be used here One-sample z-test statistics as we know about population standard deviation;

                            T.S. =  $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$  ~  N(0,1)

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = $\frac{66+68+59+67+64+66+63+62+61}{9}$

                                                    =  $\frac{576}{9}$  =  64

            $\sigma$ = population standard deviation = 3

            n = sample of values = 9

So, the test statistics  =  $\frac{64-60}{\frac{3}{\sqrt{9} } }$

                                     =  4    

The value of z-test statistics is 4.

Now at 0.05 level of significance, the z table gives a critical value of -1.96 and 1.96 for the two-tailed test.

Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean in the population is not equal to 60.