The heights of ten mails of a given locality are found to be $70, 67, 62, 68, 61, 68, 70, 64, 64, 66$ (inches). Is it reasonable to believe that the average height is $64$ inches. Test at $5 \%$ significant level assuming that for $9$ degrees of freedom is $1.833$ .

View step-by-step help

Home

Math Problems

Grade 6

Convert between customary and metric systems

Full solution

Q. The heights of ten mails of a given locality are found to be

70, 67, 62, 68, 61, 68, 70, 64, 64, 66

(inches). Is it reasonable to believe that the average height is

64

inches. Test at

5 \%

significant level assuming that for

9

degrees of freedom is

1.833

Calculate sample mean: Calculate the sample mean (average height) of the ten males. $\newline$ To find the sample mean, add up all the heights and divide by the number of heights. $\newline$ Sample mean = $(70 + 67 + 62 + 68 + 61 + 68 + 70 + 64 + 64 + 66) / 10$ $\newline$ Sample mean = $650 / 10$ $\newline$ Sample mean = $65$ inches
Calculate standard deviation: Calculate the sample standard deviation. $\newline$ First, find the differences between each height and the sample mean, square these differences, sum them up, and then divide by the number of observations minus one $(n-1)$ to get the variance. Finally, take the square root of the variance to get the standard deviation. $\newline$ Differences squared: $(70-65)^2 + (67-65)^2 + (62-65)^2 + (68-65)^2 + (61-65)^2 + (68-65)^2 + (70-65)^2 + (64-65)^2 + (64-65)^2 + (66-65)^2$ $\newline$ Differences squared: $25 + 4 + 9 + 9 + 16 + 9 + 25 + 1 + 1 + 1$ $\newline$ Sum of differences squared: $100$ $\newline$ Variance: $100 / (10-1)$ $\newline$ Variance: $100 / 9$ $\newline$ Variance $\approx 11.11$ $\newline$ Standard deviation $\approx \sqrt{11.11}$ $\newline$ Standard deviation $\approx 3.33$ inches
Perform t-test: Perform the t-test to determine if the average height is significantly different from $64$ inches. $\newline$ The t-test formula is: $t = \frac{(\text{Sample mean} - \text{Hypothesized mean})}{(\text{Standard deviation} / \sqrt{n})}$ $\newline$ Where $n$ is the sample size. $\newline$ $t = \frac{(65 - 64)}{(3.33 / \sqrt{10})}$ $\newline$ $t \approx \frac{1}{(3.33 / \sqrt{10})}$ $\newline$ $t \approx \frac{1}{(3.33 / 3.16)}$ $\newline$ $t \approx \frac{1}{1.05}$ $\newline$ $t \approx 0.95$
Compare t-values: Compare the calculated $t$ -value to the critical $t$ -value from the $t$ -distribution table. The critical $t$ -value for a $5\%$ significance level and $9$ degrees of freedom is given as $1.833$ . Since our calculated $t$ -value of $0.95$ is less than the critical $t$ -value of $1.833$ , we do not reject the null hypothesis.
Make conclusion: Make a conclusion based on the comparison of the $t$ -value and the critical value. $\newline$ Since the calculated $t$ -value is less than the critical $t$ -value, there is not enough evidence to suggest that the average height is significantly different from $64$ inches at the $5\%$ significance level.