# A parent believes the average height for 14-year-old girls differs from that of 14-year-old boys....

## Question:

A parent believes the average height for 14-year-old girls differs from that of 14-year-old boys. Estimate the difference in the height between girls and boys using a 95% confidence interval. The summary data are listed below where height is in feet.

{eq}n_1 = 40; x_1 = 5.1; s_1 = 0.2; \\ n_2 = 40; x_2 = 4.8; s_2 = 0.3 {/eq}

Based on your interval, do you think there is a significant difference between the true mean height of 14-year-old girls and boys? Explain.

## t-distribution:

The t distribution is a probability distribution. It is used to estimate the population parameters when the sample size is small and when the population variance is unknown. A t-distribution is the same as a normal distribution with a bell shape, but it has heavier tails.

## Answer and Explanation:

{eq}{n_1} = 40,{n_2} = 40,{x_1} = 5.1,{x_2} = 4.8,{s_1} = 0.2,{s_2} = 0.3 {/eq}

The formula to calculate the confidence interval for the difference between the height is as follows:

{eq}{\mu _1} - {\mu _2} = \left( {{{\bar x}_1} - {{\bar x}_2} \pm {t_{{n_1} + {n_2} - 2}}\sqrt {\frac{{{s_1}^2}}{{{n_1}}} + \frac{{{s_2}^2}}{{{n_2}}}} } \right) {/eq}

The value of {eq}{t_{{n_1} + {n_2} - 2}} {/eq} can be obtained from standard t distribution table.

Substitute the values in the above mentioned formula:

{eq}\begin{align*} {\mu _1} - {\mu _2} &= \left( {5.1 - 4.8 \pm {t_{{n_1} + {n_2} - 2}}\sqrt {\frac{{{{0.2}^2}}}{{40}} + \frac{{{{0.3}^2}}}{{40}}} } \right)\\ &= \left( {0.3 - 2.285 \times 0.057,0.3 + 2.285 \times 0.057} \right)\\ & = \left( {0.1698,0.4302} \right) \end{align*} {/eq}

Therefore, the confidence interval for the difference between the height is 0.1698, 0.4302.

The obtained confidence interval includes zero. Therefore, there is a significant difference between the true mean height of 14 years old girls and 14 year old boys.

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from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 5