Answer to Question #128703 in Statistics and Probability for wajahat

Question

A farmer is trying out a planting technique that he hopes will increase the yield on his pea plants. The
average number of pods on one of his pea plants is 145 pods with a standard deviation of 100 pods. This
year, after trying his new planting technique, he takes a random sample of 35 plants and finds the
average number of pods to be 147. He wonders whether or not this is a statistically significant increase.
i. What would the null and alternative hypotheses be for this scenario?
ii. What would the standard error be for this particular scenario?
iii. Describe in your own words how you would set the critical regions and what they would be at
an alpha level of .05.
iv. Test the null hypothesis and explain your decision

Answer 1

Solution :

Given :

$\overline{X}= 147$ ,$\sigma= 100,$ $\mu=145$ , n= 35

i)First, we develop our null and alternative hypotheses:

$H_{o}:\mu =145$

$H_{a}:\mu >145$

ii)standard error = $\frac{\sigma }{\sqrt{n}}$ =$\frac{100 }{\sqrt{35}}$ = 16.9031

iii)

Critical region:

From the alternative hypothesis, it is clear that the test is right-tailed Z test.

The level of significance is 0.05

From standard normal table or ti84 at 0.05 level of significance, the right tailed critical value is, Z= invnorm (1-0.05)= invnorm (0.95)= 1.645

iv)

The z-statistic is computed as follows:

z= $\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}$ = $\frac{147-145 }{\frac{100 }{\sqrt{35}}}$ = 0.118

Since it is observed that $z = 0.118 \le z_c = 1.64$ ,it is then concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is greater than 145, at the 0.05 significance level.