z = (153/600 – .3) / Sqrt(.3 * .7 / 600)

z = -2.41

You need a one tail test because you are looking for an inequality. Is the claim to high?

The hypothesis test is H0: p ≥ 0.30 vs H1: p < 0.30 the reason we select the hypotheses in this way is that if we have evidence to conclude the alternate hypothesis is true then you can conclude that the proportion is less than 30% and the claim is to hight. if you set up the hypothesis the other way then you'd have the H0: p ≤ 0.30 and if you fail to reject H0 then you can only say that it is plausible the 30% is to high. You can never conclude that H0 is true. This is a common misconception. the p-value for this test is P(Z < -2.41) = 0.0080 with a p-value this small we have statistical evidence that the null hypothesis is false, and the alternate hypothesis is true. The claim of 30% is to high.

First you need to find out what is your population mean, X

to test:

H0: p= 51/200

H1 p> 51/200

therefore it is a one tail test.

Test statistic:

(P-p)/([p(1-p)]/√n)

To reject Ho, Zcal must be >Z 0.95 =1.645

To test:

Zcal= (P-p)/([p(1-p)]/√n)

= [51/200-0.3]/ ([0.3(0.7)/√600]

= -0.045/(0.21/√600)

= -0.045/0.008573

= -5.248

= -5.25<1.645 Therefore Ho is true. Since Ho is true, The claim is not too high at 5% level of significance.

to find the z value, use the formula

z = (x – mean)/ standard deviation.