Question

the breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1775 pounds and a standard deviation of 80 pounds. the company believes that, due to an improvement in the manufacturing process, the mean breaking strength, μ, of the cables is now greater than 1775 pounds. to see if this is the case, 90 newly - manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1783 pounds. can we support, at the 0.10 level of significance, the claim that the population mean breaking strength of the newly - manufactured cables is greater than 1775 pounds? assume that the population standard deviation has not changed. perform a one - tailed test. then complete the parts below. carry your intermediate computations to three or more decimal places, and round your responses as specified below. (if necessary, consult a list of formulas.) (a) state the null hypothesis h₀ and the alternative hypothesis h₁. h₀: μ = 1775 h₁: μ > 1775 (b) determine the type of test statistic to use. z (c) find the value of the test statistic. (round to three or more decimal places.) 0.949 (d) find the p - value. (round to three or more decimal places.) 0.171 (e) can we support the claim that the population mean breaking strength of the newly - manufactured cables is greater than 1775 pounds? no

Explanation:

Step1: Recall z - test formula

The formula for the z - test statistic in a one - sample mean test is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size.
We are given that $\bar{x} = 1783$, $\mu=1775$, $\sigma = 80$, and $n = 90$.

Step2: Calculate the z - test statistic

Substitute the values into the formula:
\[

$$\begin{align*} z&=\frac{1783 - 1775}{\frac{80}{\sqrt{90}}}\\ &=\frac{8}{\frac{80}{\sqrt{90}}}\\ &=\frac{8\sqrt{90}}{80}\\ &=\frac{\sqrt{90}}{10}\\ &\approx\frac{9.48683}{10}\\ &= 0.949 \end{align*}$$

\]

Step3: Find the p - value

Since this is a one - tailed (right - tailed) test, the p - value is $P(Z>z)$.
Using a standard normal distribution table or calculator, for $z = 0.949$, $P(Z>0.949)=1 - P(Z\leqslant0.949)$.
From the standard normal table, $P(Z\leqslant0.949)\approx0.829$, so the p - value is $1 - 0.829=0.171$.

Step4: Make a decision

We compare the p - value with the significance level $\alpha = 0.10$.
Since the p - value ($0.171$) is greater than $\alpha(0.10)$, we fail to reject the null hypothesis.

Answer:

(a) $H_0:\mu = 1775$, $H_1:\mu>1775$
(b) $z$
(c) $0.949$
(d) $0.171$
(e) No