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An article presents a study of the failure pressures of roof panels. A sample of 12 panels constructed with 8-inch nail spacing on the intermediate framing members had a mean failure pressure [tex](\overline{X})[/tex] of 8.38 kPa with a standard deviation of 0.96 kPa. A sample of 10 panels constructed with 6-inch nail spacing on the intermediate framing members had a mean failure pressure [tex](\overline{Y})[/tex] of 9.93 kPa with a standard deviation of 1.02 kPa.

1. Can you conclude that 6-inch spacing provides a higher mean failure pressure? Find the appropriate statistic. Write your answer to three decimal places.

2. How many degrees of freedom should we use? Round your answer down to the nearest integer.

3. Can you conclude that 6-inch spacing provides a higher mean failure pressure? Find the P-value.

Answer :

A t-test suggests significant difference in mean failure pressures for different nail spacings. With a t-statistic of -4.518, the P-value is under 0.0005, indicating that 6-inch spacing likely provides higher mean failure pressure.

In this problem, we can use a t-test to determine if the difference between mean failure pressures is statistically significant. The t-test compares two means to see if they are significantly different from each other.

First, we calculate the degrees of freedom, which is the total number of observations in both groups (-2): DF = (12+10) - 2 = 20.

We then calculate the t-statistic, given by:

T = (X1 - X2) / sqrt((SD1^2 / n1) + (SD2^2 / n2))

Substituting the values provided:

T = (8.38 - 9.93) / sqrt((0.96^2 / 12) + (1.02^2 / 10)) = -4.518

Here, we have a negative t-statistic which implies the second sample mean (6-inch spacing) is larger than the first (8-inch spacing).

For the P-value, we consult a t-distribution table or software. For DF=20 and two-tailed t of -4.518, the P-value will be very much under 0.0005, telling us that a 6-inch spacing does indeed provide a significantly higher mean failure pressure.

Learn more about the topic of t-test here:

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