Answer:
The valves perform above expectations.
Explanation:
We are given the following information in the question:
Population mean, μ = 7.6 pounds per square inch
Sample size, n = 140
Sample mean,
= 7.8 pounds per square inch
Population standard deviation = 1.0 pounds per square inch
Level of significance = 0.05
We design the null and alternate hypothesis:
: μ = 7.6 pounds per square inch
: μ > 7.6 pounds per square inch
Formula:
![z_(stat) = \displaystyle\frac{\bar{x}-\mu}{(\sigma)/(√(n))}](https://img.qammunity.org/2020/formulas/mathematics/college/2baz7jnj1hsjnnxciq7fvw8atvebuldfic.png)
![z_(stat) = \displaystyle(7.8-7.6)/((1)/(√(140)))](https://img.qammunity.org/2020/formulas/mathematics/college/uhhaqm7faqx546zim24zlzh1x265qxdage.png)
![z_(stat) = 2.366](https://img.qammunity.org/2020/formulas/mathematics/college/6acpi3oxpttmn6ikja69mbbzig0s8qge0f.png)
Now, we are performing a one tail test with level of significance of 0.05, we calculate the critical value of z with the help of standard normal distribution table.
Thus,
= 1.645
Result:
Since,
![z_(stat) > z_(critical)](https://img.qammunity.org/2020/formulas/mathematics/college/nu7u1mv35whclvfbsgzokio4ops79ga24x.png)
is rejected.
Thus, we accept the alternate hypothesis.
Hence, the valve perform above expectations.