The breaking strengths of cables produced by a certain manufacturer have a mean, , of pounds, and a standard deviation of pounds. It is claimed that an improvement in the manufacturing process has increased the mean breaking strength. To evaluate this claim, newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be pounds. Assume that the population is normally distributed. Can we support, at the level of significance, the claim that the mean breaking strength has increased

The question is incomplete. The complete question is :

The breaking strengths of cables produced by a certain manufacturer have a mean of 1900 pounds, and a standard deviation of 65 pounds. It is claimed that an improvement in the manufacturing process has increased the mean breaking strength. To evaluate this claim, 150 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1902 pounds. Assume that the population is normally distributed. Can we support, at the 0.01 level of significance, the claim that the mean breaking strength has increased?

Solution :

Given data :

Mean, μ = 1900

Standard deviation, σ = 65

Sample size, n = 150

Sample mean, = 1902

Level of significance = 0.01

The hypothesis are :

Test statics :

We use the z test as the sample size is large and we know the population standard deviation.

Finding the p-value:

P-value = P(Z > z)

             = P(Z > 0.38)

             = 1 - P(Z < 0.38)

From the z table. we get

P(Z < 0.38) = 0.6480

Therefore,

P-value = 1 - P(Z < 0.38)

            = 1 - 0.6480

             = 0.3520

Decision :

If the p value is less than 0.01, then we reject the , otherwise we fail to reject  .

Since the value of p = 0.3520 > 0.01, the level of significance, then we fail to reject  .

Conclusion :

At a significance level of 0.01, we have no sufficient evidence to support that the mean breaking strength has increased.

78.59 is the mean

step-by-step explanation: