Big babies: The National Health Statistics Reports described a study in which a sample of 315 one-year-old baby boys were weighed. Their mean weight was 25.6 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys is greater than 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the =α0.05 level of significance and the critical value method with the

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otaku20961 otaku20961

03-05-2021
Mathematics

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Big babies: The National Health Statistics Reports described a study in which a sample of 315 one-year-old baby boys were weighed. Their mean weight was 25.6 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys is greater than 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the =α0.05 level of significance and the critical value method with the

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joaobezerra joaobezerra · Answer 1 · 2021-05-06T20:58:51-04:00

Answer:

The pvalue of the test is 0.0012 < 0.05, which means that the data provides convincing evidence that the pediatrician's claim is true.

Step-by-step explanation:

A pediatrician claims that the mean weight of one-year-old boys is greater than 25 pounds.

This means that at the null hypothesis, we test that the mean is 25 pounds, that is:

[tex]H_0: \mu = 25[/tex]

At the alternate hypothesis, we test that it is more than 25 pounds, that is:

[tex]H_a: \mu > 25[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

25 is tested at the null hypothesis:

This means that [tex]\mu = 25[/tex]

The National Health Statistics Reports described a study in which a sample of 315 one-year-old baby boys were weighed. Their mean weight was 25.6 pounds with standard deviation 5.3 pounds.

This means that [tex]n = 315, \mu = 25.6, \sigma = 5.3[/tex]

Value of the test-statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{25.6 - 25}{\frac{5.3}{\sqrt{315}}}[/tex]

[tex]z = 3.04[/tex]

Pvalue of the test and decision:

The pvalue of the test is the probability of finding a mean above 25.6 pounds, which is 1 subtracred by the pvalue of z = 3.04.

Looking at the z-table, z = 3.04 has a pvalue of 0.9988

1 - 0.9988 = 0.0012

The pvalue of the test is 0.0012 < 0.05, which means that the data provides convincing evidence that the pediatrician's claim is true.

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