Respuesta :
Answer:
There is sufficient evidence to conclude that child booster seats meet the specific requirement.
Step-by-step explanation:
Sample: 697, 759, 1266, 621, 569, 432
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{4344}{6} = 724[/tex]
Sum of squares of differences = 415616
[tex]S.D = \sqrt{\frac{415616}{5}} = 288.31[/tex]
We are given the following in the question:
Population mean, μ = 1000 hic
Sample mean, [tex]\bar{x}[/tex] = 724
Sample size, n = 16
Alpha, α = 0.05
Sample standard deviation, s = 288.31
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 1000\text{ hic}\\H_A: \mu < 1000\text{ hic}[/tex]
We use one-tailed(left) t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{724 - 1000}{\frac{288.31}{\sqrt{6}} } = -2.344[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 5 degree of freedom } = -2.015[/tex]
Calculation the p-value from table,
P-value = 0.033
Since,
Since, the p value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
We conclude that the measurement is less than 1000 hic.
Thus, there is sufficient evidence to conclude that child booster seats meet the specific requirement.