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Answer :
To find the p-value for this hypothesis test, we can use the t-test since the sample size is small (n = 7) and the population standard deviation is unknown. In this case, the test statistic can be calculated using the formula for the one-sample t-test:
[tex]t = \frac{\bar{x} - \mu}{s/\sqrt{n}}[/tex]
where:
- [tex]\bar{x}[/tex] is the sample mean,
- [tex]\mu[/tex] is the population mean (101.4 in this case),
- [tex]s[/tex] is the sample standard deviation (6.3 as stated),
- [tex]n[/tex] is the sample size (7).
a) If the observed mean [tex]\bar{x} = 95.9[/tex]:
Calculate the t-value:
[tex]t = \frac{95.9 - 101.4}{6.3/\sqrt{7}} = \frac{-5.5}{2.379} \approx -2.31[/tex]
Determine the degrees of freedom (df):
[tex]df = n - 1 = 7 - 1 = 6[/tex]
Using a t-table or calculator, check the two-tailed p-value for [tex]t = -2.31[/tex] with 6 degrees of freedom.
The p-value [tex]approx 0.060[/tex].
b) If the observed mean [tex]\bar{x} = 99.8[/tex]:
Calculate the t-value:
[tex]t = \frac{99.8 - 101.4}{6.3/\sqrt{7}} = \frac{-1.6}{2.379} \approx -0.67[/tex]
Use the degrees of freedom (df = 6).
Check the two-tailed p-value for [tex]t = -0.67[/tex] with 6 degrees of freedom.
The p-value [tex]approx 0.529[/tex].
c) If the observed mean [tex]\bar{x} = 103.3[/tex]:
Calculate the t-value:
[tex]t = \frac{103.3 - 101.4}{6.3/\sqrt{7}} = \frac{1.9}{2.379} \approx 0.80[/tex]
Again, use the same degrees of freedom (df = 6).
Check the two-tailed p-value for [tex]t = 0.80[/tex] with 6 degrees of freedom.
The p-value [tex]approx 0.455[/tex].
In summary, the p-values for each sample mean are approximately 0.060, 0.529, and 0.455, respectively. These p-values help us determine how likely it is to observe a sample mean as extreme as the one calculated, assuming that the null hypothesis is true.
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