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Answer :
Answer:
Question 1
The test statistics is [tex]t = 0.44[/tex]
The decision rule is
Fail to reject the null hypothesis
The conclusion
There is no sufficient evidence to show that there is a difference in the average time to complete the installation of the solar panels
Question 2
The degree of freedom is [tex]df = 92 [/tex]
The decision rule is
Reject the null hypothesis
The conclusion
There is sufficient evidence to show that there is a difference in the average time to complete the installation of the solar panels
Explanation:
Considering Question 1
Here we are told to provide the test statistics
From the question we are told that
The first sample size is [tex]n_1 = 31[/tex]
The second sample size is [tex]n_ 2 = 46[/tex]
The first sample mean is [tex]\= x_1 = 525 \ minutes[/tex]
The first standard deviation is [tex]\sigma_1 = 47.7[/tex]
The second sample mean is [tex]\= x_2 = 520 \ minutes[/tex]
The second standard deviation is [tex]\sigma_2 = 48.2[/tex]
The null hypothesis is [tex]H_o : \mu_1 - \mu_2 = 0[/tex]
The alternative hypothesis is [tex]H_a : \mu_1 - \mu_2 \ne 0[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = n_1 + n_2 -2[/tex]
=> [tex]df = 31 + 46 -2[/tex]
=> [tex]df = 75 [/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{(\= x_1 - \= x_2 )-0}{ \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} } }[/tex]
=> [tex]t = \frac{( 525- 520 )-0}{ \sqrt{\frac{47.7^2}{31} + \frac{48.2^2}{46} } }[/tex]
=> [tex]t = 0.44[/tex]
Let assume that the level of confidence is [tex]\alpha = 0.05[/tex]
Generally the probability of t at a degree of freedom of is [tex]df = 75[/tex]
[tex]P(t > 0.44 ) = 0.33060124 [/tex]
Generally the p-value is mathematically represented as
[tex]p-value = 2 * P(t > 2.398)[/tex]
=> [tex]p-value = 2 * 0.33060124[/tex]
=> [tex]p-value = 0.66120[/tex]
From the value obtain we see that [tex]p-value > \alpha[/tex] hence we fail to reject the null hypothesis
The conclusion is that there is no sufficient evidence to show that there is a difference in the average time to complete the installation of the solar panels
Considering Question 2
Here we are told to provide the degree of freedom
From the question we are told
The first sample size is [tex]n_1 = 39[/tex]
The first sample mean is [tex]\= x_1 = 582 \ minutes[/tex]
The first standard deviation is [tex]\sigma_2 = 63.8[/tex]
The second sample size is [tex]n_ 2 = 55[/tex]
The second sample mean is [tex]\= x_2 = 542 \ minutes[/tex]
The second standard deviation is [tex]\sigma_2 = 97.8 [/tex]
The null hypothesis is [tex]H_o : \mu_1 - \mu_2 = 0[/tex]
The alternative hypothesis is [tex]H_a : \mu_1 - \mu_2 \ne 0[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = n_1 + n_2 -2[/tex]
=> [tex]df = 39 + 55 -2[/tex]
=> [tex]df = 92 [/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{(\= x_1 - \= x_2 )-0}{ \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} } }[/tex]
=> [tex]t = \frac{( 582 - 542 )-0}{ \sqrt{\frac{63.8^2}{39} + \frac{97.8^2}{55} } }[/tex]
=> [tex]t = 2.398[/tex]
Let assume that the level of confidence is [tex]\alpha = 0.05[/tex]
Generally the probability of t at a degree of freedom of is [tex]df = 92 [/tex]
[tex]P(t > 2.398 ) = 0.00925214[/tex]
Generally the p-value is mathematically represented as
[tex]p-value = 2 * P(t > 2.398)[/tex]
=> [tex]p-value = 2 * 0.00925214[/tex]
=> [tex]p-value = 0.0185[/tex]
From the value obtain we see that [tex]p-value < \alpha[/tex] hence we reject the null hypothesis
The conclusion is that there is sufficient evidence to show that there is a difference in the average time to complete the installation of the solar panels
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