An engineer who is studying the tensile strength of a steel alloy intended for use in golf club shafts knows that tensile strength is approximately normally distributed with σ=75 psi. a random sample of 12 specimens has a sample mean tensile strength of 3450 psi. is there evidence to support the claim that mean strength is greater than 3400 psi. state and test the hypothesis at significance level α =0.01. (2 points)

We conclude that the mean strength is equal to 3400 psi.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 3400 psi

Sample mean, = 3450 psi

Sample size, n = 12

Alpha, α = 0.051

Population standard deviation, σ =75 psi

First, we design the null and the alternate hypothesis

We use one-tailed(right) z test to perform this hypothesis.

Formula:

Putting all the values, we have

Now,

Since,  

We fail to reject the null hypothesis and accept the null hypothesis. Thus, we conclude that the mean strength is equal to 3400 psi.

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step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.