Question

An article presents a study of the failure pressures of roof panels. A sample of 15 panels constructed with 8-inch nail spacing on the intermediate framing members had a mean failure pressure of 8.48 kPa with a standard deviation of 0.96 kPa. A sample of 15 panels constructed with 6-inch nail spacing on the intermediate framing members had a mean failure pressure of 9.93 kPa with a standard deviation of 1.02 kPa. Can you conclude that 6-inch spacing provides a higher mean failure pressure

Answer 1

Answer:

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance

Step-by-step explanation:

From the question we are told that:

Sample size 1 $n_1=15$

Spacing 1 $s_1=8$

Mean 1 $\=x_1=8.48$

Standard deviation 1 $\sigma_1=0.96 kPa$

Sample size 2 $n_2=16$

Spacing 2 $s_2=6$

Mean 1 $\=x_1=9.93$

Standard deviation 1$\sigma_1=1.02 kPa$

Generally the hypothesis is mathematically given as

Null $H_0:\mu_1-\mu_2$

Alternative $H_a:\mu_1\mu_2$

Generally the equation for pooled estimate is mathematically given by

  $S=\sqrt{\frac{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2}{n_1+n_2-2} }$

Therefore

  $S=\sqrt{\frac{(15-1)0.96^2+(16-1)1.02^2}{15+15-2} }$

 $S=0.9905$

Generally the equation for test statistics is mathematically given by

 $t=\frac{\=x_1+\=x_2}{S\sqrt{\frac{1}{n _1}+\frac{1}{n_2}}}$

 $t=\frac{\=8.48-9.93}{0.9905\sqrt{\frac{1}{15}+\frac{1}{15}}}$

 $t=-4.00091$

Therefore From table

 $P value=P(t_{15+15-2}=4.00091)$

 $P value=0.00041918$

 $P$

Therefore

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance