1. Mapalad Integrated High School determined students’Body Mass Index (BMI) at the opening of classes. It hasbeen recorded that the average height of female students is154.2 centimeters with a standard deviation of 9centimeters. The researcher conducted her own study andshe randomly selected 40 female students. In her study,she got an average of 156.7 centimeters. Is there a reasonto believe the claims of the school? Use 5% level ofsignificance in testing the hypothesis.

To answer this question, we will conduct a hypothesis test.

Step 1: State the null hypothesis (H0) and the alternative hypothesis (H1). H0: μ = 154.2 (The school's claim is correct) H1: μ ≠ 154.2 (The school's claim is incorrect)

Step 2: Identify a test statistic. Since we know the standard deviation and we have a large sample size (n>30), we can use the z-score formula.

Step 3: Calculate the test statistic. Z = (X - μ) / (σ/√n) Where: X = sample mean = 156.7 μ = population mean = 154.2 σ = standard deviation = 9 n = number of observations = 40

Z = (156.7 - 154.2) / (9/√40) = 1.39

Step 4: Determine the critical value for 5% level of significance. The critical value of z for a two-tailed test at 5% level of significance is ±1.96.

Step 5: Make a decision. Since the calculated z-score (1.39) is less than the critical z-score (1.96), we fail to reject the null hypothesis.

Conclusion: There is not enough evidence at the 5% level of significance to reject the school's claim that the average height of female students is 154.2 cm.

To answer this question, we will conduct a hypothesis test. Here are the steps:

Step 1: State the null and alternative hypotheses. The null hypothesis (H0) is that the average height of female students is 154.2 cm. The alternative hypothesis (H1) is that the average height of female students is not 154.2 cm.

Step 2: Identify a test statistic. Since we know the standard deviation and we have a large sample size (n>30), we can use the z-test.

Step 3: Compute the test statistic. Z = (X - μ) / (σ/√n) Where: X = sample mean = 156.7 cm μ = population mean = 154.2 cm σ = standard deviation = 9 cm n = sample size = 40

Z = (156.7 - 154.2) / (9/√40) = 1.39

Step 4: Determine the P-value. The P-value associated with a 1.39 z-score is 0.0823.

Step 5: Compare the P-value to the significance level. The P-value (0.0823) is greater than the 5% level of significance.

Step 6: Make a decision. Since the P-value is greater than the level of significance, we fail to reject the null hypothesis.

Conclusion: There is not enough evidence to reject the school's claim that the average height of female students is 154.2 cm.