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Answer :
Final answer:
To determine if the 6-inch spacing provides a higher mean failure pressure, a hypothesis test can be used. The two-sample t-test with unequal variances can determine if there is a significant difference in the mean failure pressures. The test statistic is calculated using the sample means, standard deviations, and sample sizes.
Explanation:
To determine whether the 6-inch nail spacing provides a higher mean failure pressure, we can compare the means of the two samples. We will use a hypothesis test to draw a conclusion. Let's assume the null hypothesis (H0) is that there is no difference in mean failure pressure between the two nail spacing. The alternative hypothesis (Ha) is that the mean failure pressure with 6-inch spacing is higher. We will use a two-sample t-test with unequal variances to test this.
The test statistic for the two-sample t-test is calculated using the formula:
t = (mean1 - mean2) / sqrt((s1^2/n1) + (s2^2/n2))
where mean1 and mean2 are the sample means, s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes. We can calculate the test statistic and compare it to the critical value for the desired level of significance to determine if there is a significant difference.
Learn more about Two-sample t-test here:
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Answer:
There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches
Reject [tex]H_0\ at\ 5%[/tex] level of significance
Step-by-step explanation:
From the question we are told that:
Sample size 1 [tex]n_1=15[/tex]
Spacing 1 [tex]s_1=8[/tex]
Mean 1 [tex]\=x_1=8.48[/tex]
Standard deviation 1 [tex]\sigma_1=0.96 kPa[/tex]
Sample size 2 [tex]n_2=16[/tex]
Spacing 2 [tex]s_2=6[/tex]
Mean 1 [tex]\=x_1=9.93[/tex]
Standard deviation 1[tex]\sigma_1=1.02 kPa[/tex]
Generally the hypothesis is mathematically given as
Null [tex]H_0:\mu_1-\mu_2[/tex]
Alternative [tex]H_a:\mu_1\mu_2[/tex]
Generally the equation for pooled estimate is mathematically given by
[tex]S=\sqrt{\frac{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2}{n_1+n_2-2} }[/tex]
Therefore
[tex]S=\sqrt{\frac{(15-1)0.96^2+(16-1)1.02^2}{15+15-2} }[/tex]
[tex]S=0.9905[/tex]
Generally the equation for test statistics is mathematically given by
[tex]t=\frac{\=x_1+\=x_2}{S\sqrt{\frac{1}{n _1}+\frac{1}{n_2}}}[/tex]
[tex]t=\frac{\=8.48-9.93}{0.9905\sqrt{\frac{1}{15}+\frac{1}{15}}}[/tex]
[tex]t=-4.00091[/tex]
Therefore From table
[tex]P value=P(t_{15+15-2}=4.00091)[/tex]
[tex]P value=0.00041918[/tex]
[tex]P<0.0005[/tex]
Therefore
There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches
Reject [tex]H_0\ at\ 5%[/tex] level of significance
