Question

A researcher wants to check the claim that convicted burglars spend less than 18.7 months in jail, on average. She takes a random sample of 67 such cases from court files and finds that the mean is 16.8 months with a standard deviation of 7.3 months.
a. Is there enough evidence to support the claim using a level of significance of 1%?
b. Enter the Null Hypothesis for this test: H0:_________
c. Enter the Alternative Hypothesis for this test: H1: __________d. What is the p-value for this hypothesis test? Round your answer to four decimal places.e. What is the decision based on the given sample statistics?

Answer 1

Answer:

a. No, there is not enough evidence to support the claim using a level of significance of 1%.

b. Null Hypothesis, $H_0$ : $\mu \geq$ 18.7 months  

c. Alternate Hypothesis, $H_A$ : $\mu$ < 18.7 months

d. P-value of hypothesis test = 0.0199 or 1.99%

e. We conclude that the convicted burglars spend more than or equal to 18.7 months in jail, on average.

Step-by-step explanation:

We are given that a researcher wants to check the claim that convicted burglars spend less than 18.7 months in jail, on average.

She takes a random sample of 67 such cases from court files and finds that the mean is 16.8 months with a standard deviation of 7.3 months.

Let $\mu$ = true mean time convicted burglars spend in jail.

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 18.7 months   {means that the convicted burglars spend more than or equal to 18.7 months in jail, on average}

Alternate Hypothesis, $H_A$ : $\mu$ < 18.7 months   {means that the convicted burglars spend less than 18.7 months in jail, on average}

The test statistics that will be used here is One-sample t test statistics as 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 time spent in jail = 16.8 months

             s = sample standard deviation = 7.3 months

             n = sample of cases = 67

So, test statistics  =   $\frac{16.8-18.7}{\frac{7.3}{\sqrt{67} } }$  ~ $t_6_6$

                               =  -2.13

Hence, the value of test statistics is -2.13.

Now, P-value of the test statistics is given by;

      P-value = P( $t_6_6$ < -2.13) = 0.0199 or 1.99%

• If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

• If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.0199 or 1.99% which is clearly higher than the level of significance of 1%, so we will not reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the convicted burglars spend more than or equal to 18.7 months in jail, on average.