Question

A school administrator claims that less than 50% of the students of the school are dissatisfied

by the community cafeteria service. Test this claim by using sample data obtained from a survey of

500 students of the school where 54% indicated their dissatisfaction of the community cafeteria

service. Use 𝐠= 0.05.

1) Complete the table.

n X p₀ z p-value

a) 120 21 5%

b) 138 32 7%

c) 200 45 10%

d) 392 102 18%

e) 612 236 20%

f) 100 40 8%

g) 248 51 10%

h) 312 100 12%

Answer 1

When the sample size is large the sample proportion is normally distributed.

$$np=500(0.5)=250>30$$ $$nq=500(0.5)=250>30$$

With $n=500,$ the Central Limit Theorem applies.

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p\ge0.50$

$H_a:p<0.50$

This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.

Evidence:

Based on the information provided, the significance level is $\alpha = 0.05 ,$ and the critical value for a left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z: z < -1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.54-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{500}}}=3.5777$$

Since it is observed that $z =3.5777 \ge-1.6449= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z<3.5777)= 0.999827,$ and since $p=0.999827>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is less than 0.50, at the $\alpha = 0.05$ significance level.

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} n & X & p_0 & z & p-value \\ \hline 120 & 21 & 0.05 & 6.2828 & 0 \\ \hdashline 138 & 32 & 0.07 & 7.4534 & 0\\ \hdashline 200 & 45 & 0.10 & 5.8926 & 0\\ \hdashline 392 & 102 & 0.18 & 4.1333 &0.000036 \\ \hdashline 612 & 236 & 0.20 & 11.4800 & 0\\ \hdashline 100 & 40 & 0.08 & 11.7954& 0\\ \hdashline 248 & 51 & 0.10 & 5.5457&0 \\ \hdashline 312 & 100 & 0.12 &10.8990 &0 \\ \end{array}$$