Question

Fuses produced by a company will blow in 12.40 minutes as the average when
overloaded. Suppose the mean blow time of 20 fuses subjected to overload is
10.63 minutes and S.D 2.48 mts. Does this information tend to support or
refuse the claim that the population mean blow time is 12.40 mts ?​

Answer 1

This information tend to  refuse the claim that the population mean blow time is 12.40 minutes.

Step-by-step explanation:

We are given that Fuses produced by a company will blow in 12.40 minutes as the average when  overloaded. Suppose the mean blow time of 20 fuses subjected to overload is  10.63 minutes and S.D 2.48 minutes.

We have to test the claim that the population mean blow time is 12.40 minutes or not.

Let $\mu$ = population mean blow time

So, Null Hypothesis, $H_0$ : $\mu$ = 12.40 minutes  {means that the population mean blow time is 12.40 minutes}

Alternate Hypothesis, $H_a$ : $\mu\neq$ 12.40 minutes  {means that the population mean blow time is different from 12.40 minutes}

The test statistics that will be used here is t-test statistics because we don't know about the population standard deviation.

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

where,  $\bar X$ = sample mean blow time = 10.63 minutes

              s = sample standard deviation = 2.48 minutes

              n = sample of fuses = 20

So, Test statistics =  $\frac{10.63-12.40}{\frac{2.48}{\sqrt{20} } }$  ~ $t_1_9$

                               =  -3.192

Since we are not given the level of significance in the question so we assume it to be 5%. At 5% significance level, the t table gives critical values between -2.093 and 2.093 at 19 degree of freedom. Since our test statistics does not lie within the ranges of critical values so we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the population mean blow time is different from 12.40 minutes.