The average height of females in the freshman class of a certain college has historically been 162.5 centimeters with a standard

Question

7 days ago

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The average height of females in the freshman class of a certain college has historically been 162.5 centimeters with a standard

deviation of 6.9 centimeters. is there reason to believe that there has been a change in the average height if a random sample of 50 females in the present freshman class has an average height of 165.2 centimeters? use a p-value in your conclusion. assume the standard deviation remains the same.

Mathematics

2 answers:

AnnZ [3.9K] · Answer 1 · 2026-02-26T15:07:07+03:00

Answer:

Step-by-step explanation:

Let X represent the average height of females in the freshman class.

X is distributed as N(162.5, 6.9)

Sample mean is 165.2

Sample size equals 50

Given that the population standard deviation is known and n > 30, we can utilize the z statistic.

Z = (x bar-mu)/Sigma/sqrt n

Thus, Z = $\frac{165.2-162.5}{6.9/\sqrt{50} }$

= 2.766

Assume alpha is 5%

The critical z-value for a confidence level of 95% is 1.96

Our calculated z-value exceeds 1.96

The p-value = 0.001, which is less than 0.05 (our alpha)

This leads to the conclusion that we reject the null hypothesis.

The sample mean is higher than the population mean.

This supports the alternative hypothesis as valid.

Svet_ta [4.3K] · Answer 2 · 2026-02-26T15:06:15+03:00

To determine if there is evidence suggesting a change in average height, we can conduct a right-tailed test and formulate both null and alternative hypotheses.

H₀ (null hypothesis): μ = 162.5
H₁ (alternative hypothesis): μ > 162.5

With two samples to analyze, we can calculate the z-score using the formula provided below.

$Z = \frac{(\chi - \mu)}{ \sigma / \sqrt{n}}$

In this formula, Z symbolizes the z-score, Χ denotes the new sample mean, μ indicates the theoretical average, δ represents the standard deviation, and n signifies the sample size. Based on the gathered values,

$Z = \frac{(165.2 - 162.5)}{6.9 / \sqrt{50}}$
$Z \approx 2.77$

Assuming a significance level of α = 0.05. With a z-score of 2.77, we can reference the z-table to ascertain the p-value. This yields P(Z > 2.77) =.0028. Since our p-value is below α, we reject the null hypothesis, indicating that the average height of female freshman students has indeed shifted.

The average height of females in the freshman class of a certain college has historically been 162.5 centimeters with a standard

The constant pay of $25 signifies the intercept