Answer to Question #344797 in Statistics and Probability for jdee

Question

The leader of the jeepney drivers’ association claims that the average daily take-home pay of all jeepney drivers in Pasay city is Php 400.00. A random sample of 100 jeepney drivers in Pasay city was interviewed and the average daily take-home pay of these drivers is found to be Php 425. Use a 0.05 level of significance and assume the standard deviation is Php 92.

 Ծ = 92          Ẋ = 425           ﬠ = 400           n = 100         level of significance = 0.05

 

1. State the hypothesis

  Ho: u = 400

  Ha: u ‡ 400

 

2. Type of test: Two-tailed test

 

 

3. Critical Value

 

 

4. Test statistics

 

 

5. Reject or accept the null hypothesis

Answer 1

1. State the hypothesis

H₀: $\mu=400$

H_a: $\mu\ne400$

2. Type of test: Two-tailed test

 

 

3. Critical Value

 $\alpha=0.05, \alpha/2=0.025, z_{0.025}=1.96$ (using t-table)

 So, the rejection region for this two-tailed test is $R=\left\{z:∣z∣>1.96\right\}.$

4. Test statistics

$z=\frac{\overline{x}-\mu}{\sigma/\sqrt{n}}=\frac{425-400}{92/\sqrt{100}}=2.72$

5. Reject or accept the null hypothesis

Since 1.96<2.72, $z_{0.025}<z$ . So z is at rejection region.

Thus, $H_0$ should be rejected. The average daily take home pay of all jeepney drivers in Pasay City is different from P400.00