A garden center wants to store leftover packets of veg- etable seeds for sale the following spring, but the center is con- cerned that the seeds may not germinate at the same rate a year later. The manager finds a packet of last year’s green bean seeds and plants them as a test. Although the packet claims a germi- nation rate of 92%, only 171 of 200 test seeds sprout. Is this evidence that the seeds have lost viability during a year in stor- age? Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.
(a) Verify the three conditions for using the central limit theorem for inference onπ.
(b) Conduct a significance test to decide if there is sufficient evidence that the seeds have lost viability during a year in storage.

the above claim shows that either it is null hypothesis or it could be alternative hypothesis. The null hypothesis shows that the proportion is equal to the value mentioned in the claim. if null hypothesis claim is true, then alternative hypothesis states the opposite of the null hypothesis.

[tex]H_{O}[/tex] : p = 92 percent = 0.92

[tex]H_{a}[/tex] : p ∠ 0.92

CONDITION:

one-proportion z test will be used to test the claim about one population proportion.

Conditions:

Following condition will be followed:

Random sample
10 percent condition
Success failure condition

Random sample:

This condition is Satisfied, assuming that the sample is a random sample.

10 percent condition:

This condition is also Satisfied, since the 200 seed sprouts is less than 10 percent of all seed sprouts.

Success-failure condition:

Satisfied, since n[tex]P_{o}[/tex] and n(l -[tex]P_{o}[/tex] ) are both at least 10.

n[tex]P_{o}[/tex] = 200(0.92) = 184 [tex]\geq[/tex] 10

n(l -[tex]P_{o}[/tex] ) = 200(1 - 0.92) = 16 [tex]\geq[/tex] 10

All conditions are satisfied, assuming that the sample is a random sample .

HYPOTHESIS TEST :

The sample proportion is the number of successes divided by the sample size:

p=[tex]\frac{x}{n}[/tex]=[tex]\frac{171}{200}[/tex]

=0.855

Determine the value of the test statistic

z=[tex]\frac{p-p_{o} }{\sqrt{p_{o}(1-p_{o}})/n }[/tex]

putting the value in eq 1

z=-3.39

so the P-value is the probability to find out the value of the test statistic, if null hypothesis is true then we have to find the P-value using the normal probability table in the appendix.

P = P( Z ∠ -3.39) = 0.0003

If the P-value is smaller than the significance level [tex]\alpha[/tex], then the null hypothesis will be rejected:

P ∠ 0.05 ==> reject [tex]H_{O}[/tex]

There is sufficient evidence to support the claim that the seeds have lost viability during a year in storage .