Answer :
The appropriate conclusions are as follows:
a ----> c
b ----->c
c ------>c
d ------>a
What is the Null Hypothesis?
The null hypothesis is an assertion that the sample observations are the result of chance. The claim is that the surveyors made it because they wanted to look at the data.
From the question, we know that,
The population mean is μ=0.5 in
Usually, the Null hypothesis is H0 : μ=0.5 in
The alternative hypothesis is H0 : μ [tex]\neq[/tex] 0.5 in
Considering the parameter given for part a
The sample size is n = 15
The test statistics is t = 1.66
The level of significance α=0.05
The degree of freedom would be evaluated as follows
df= n-1
df=15-1=14
Now using the critical value formula, we get [tex]t_{\frac{a}{2} } df[/tex] =2.145
Looking at the value of t and [tex]t_{\frac{a}{2} }df[/tex] we see that [tex]t < t_{\frac{a}{2} }[/tex] so the null hypothesis would not be rejected.
Considering the parameter given for part b,
The sample size is n = 15
The test statistics is t = 1.66
The level of significance α=0.05
The degree of freedom would be evaluated as follows
df= n-1
df=15-1=14
Now using the critical value formula, we get [tex]t_{\frac{a}{2} } df[/tex]= -2.145
Looking at the value of t, we see that t does not lie in the area covered , so the null hypothesis would not be rejected.
Considering the parameter given for part c,
The sample size is n = 26
The test statistics is t = -2.55
The level of significance α=0.01
The degree of freedom would be evaluated as follows
df= n-1
df=26-1=25
Now using the critical value formula, we get [tex]t_{\frac{a}{2} } df[/tex]= 2.787
Looking at the value of t, we see that t does not lie in the area covered , so the null hypothesis would not be rejected.
Considering the parameter given for part d,
The sample size is n = 26
The test statistics is t = -2.55
The level of significance α=0.01
The degree of freedom would be evaluated as follows
df= n-1
df=26-1=25
Now using the critical value formula, we get [tex]t_{\frac{a}{2} } df[/tex]= -2.787
Looking at the value of t, we see that t lies in the area covered by , so the null hypothesis would be rejected.
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