Question

A certain manufactured product is supposed to contain 23% potassium by weight. A sample of 10 specimens of this product had an average percentage of 23.2 with a standard deviation of 0.2. If the mean percentage is found to differ from 23, the manufacturing process will be recalibrated.
a. State the appropriate null and alternate hypotheses.
b. Should the process be recalibrated? Explain.
c. Compute the P-value.

Answer 1

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 23%    

    Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 23%

(b) We conclude that the mean percentage is  different from 23 and the manufacturing process will be re-calibrated.

(c) P-value is 0.6%.

Step-by-step explanation:

We are given that a certain manufactured product is supposed to contain 23% potassium by weight.

A sample of 10 specimens of this product had an average percentage of 23.2 with a standard deviation of 0.2.

Let $\mu$ = mean percentage of potassium by weight.

(a) Null Hypothesis, $H_0$ : $\mu$ = 23%      {means that the mean percentage is equal to 23 and the manufacturing process will not be re-calibrated}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 23%     {means that the mean percentage is  different from 23 and the manufacturing process will be re-calibrated}

The test statistics that would be used here One-sample t-test statistics as we don't know about population standard deviation;

                                T.S. =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean percentage = 23.2

            s = sample standard deviation = 0.2

            n = sample of specimens = 10

So, the test statistics  =  $\frac{23.2-23}{\frac{0.2}{\sqrt{10} } }$  ~ $t_9$

                                     =  3.162

The value of t test statistic is 3.162.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of -2.262 and 2.262 at 9 degree of freedom for two-tailed test.

(b) Since our test statistic doesn't lie within the range of critical values of t, 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 percentage is  different from 23 and the manufacturing process will be re-calibrated.

(c) The P-value of the test statistics is given by;

            P-value = P( $t_9$ > 3.162) = 0.006 or 0.6%