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Answer:
We conclude that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours.
Step-by-step explanation:
We are given that a sports statistician claim that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours.
The mean winning time of a sample of 30 randomly selected Boston marathon women's open division champions is 2.60 hours. assume the population standard deviation is 0.32 hours.
Let [tex]\mu[/tex] = mean winning times for Boston marathon women's open division champions.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu\geq[/tex] 2.68 hours    {means that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours}
Alternate Hypothesis, [tex]H_A[/tex] : p < 2.68 hours    {means that the mean winning times for Boston marathon women's open division champions is less than 2.68 hours}
The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;
             T.S. =  [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean winning time = 2.60 hours
      [tex]\sigma[/tex] = population standard deviation = 0.32 hours
       n = sample of women's open division champions = 30
So, test statistics  =  [tex]\frac{2.60-2.68}{\frac{0.32}{\sqrt{30} } }[/tex]
                =  -1.37
The value of z test statistics is -1.37.
Now, P-value of the test statistics is given by following formula;
     P-value = P(Z < -1.37) = 1 - P(Z [tex]\leq[/tex] 1.37)
            = 1 - 0.9147 = 0.0853
Since, P-value of the test statistics is more than the level of significance as 0.0853 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.
Therefore, we conclude that the mean winning times for Boston marathon women's open division champions is at least 2.68 hours.
Using the z-distribution, it is found that since the p-value of the test is of 0.0853 > 0.05, you cannot reject the claim.
At the null hypothesis, it is tested if the mean winning time is of at least 2.68 hours, that is:
[tex]H_0: \mu \geq 2.68[/tex]
At the alternative hypothesis, it is tested if it is less than 2.68 hours, that is:
[tex]H_1: \mu < 2.68[/tex]
We have the standard deviation for the population, hence, the z-distribution is used to solve this question.
The test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- [tex]\sigma[/tex] is the standard deviation of the sample.
- n is the sample size.
In this problem, the values of the parameters are as follows:
[tex]\overline{x} = 2.6, \mu = 2.68, \sigma = 0.32, n = 30[/tex]
Hence, the value of the test statistic is:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{2.6 - 2.68}{\frac{0.32}{\sqrt{30}}}[/tex]
[tex]z = -1.37[/tex]
Using a calculator, the p-value for the test is of 0.0853.
Since the p-value of the test is of 0.0853 > 0.05, you cannot reject the claim.
You can learn more about the use of the z-distribution to test an hypothesis at https://brainly.com/question/16313918
