Question

assume that a simple random sample has been selected from a normally distributed population and test the given claim. identify the null and alternative hypotheses, test statistic, p - value, and state the final conclusion that addresses the original claim
a safety administration conducted crash tests of child booster seats for cars. listed below are results from those tests, with the measurements given in hic (standard head injury condition units). the safety requirement is that the hic measurement should be less than 1000 hic. use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 1000 hic do the results suggest that all of the child booster seats meet the specified requirement?
620 646 1024 579 520 571
what are the hypotheses?
a ( h_{0}:mu < 1000 ) hic
( h_{1}:mugeq1000 ) hic
b ( h_{0}:mu = 1000 ) hic
( h_{1}:mu < 1000 ) hic
c ( h_{0}:mu > 1000 ) hic
( h_{1}:mu < 1000 ) hic
d ( h_{0}:mu = 1000 ) hic
( h_{1}:mugeq1000 ) hic
identify the test statistic.
( t=square ) (round to three decimal places as needed.)

Explanation:

Step1: Determine the null and alternative hypotheses

The null hypothesis \(H_0\) is a statement of equality. The claim is that the mean is less than \(1000\) hic. So, the null hypothesis \(H_0:\mu = 1000\) hic (the status - quo, no difference from the hypothesized value) and the alternative hypothesis \(H_1:\mu<1000\) hic (the claim we are trying to find evidence for).

For the test statistic:
First, calculate the sample mean \(\bar{x}\) and sample standard deviation \(s\).
The sample data \(x=\{620,646,1024,579,520,571\}\)
The sample size \(n = 6\)
\(\bar{x}=\frac{620 + 646+1024+579+520+571}{6}=\frac{3960}{6}=660\)
The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\)
\(\sum_{i=1}^{6}(x_i - 660)^2=(620 - 660)^2+(646-660)^2+(1024 - 660)^2+(579-660)^2+(520 - 660)^2+(571-660)^2\)
\(=(- 40)^2+(-14)^2+(364)^2+(-81)^2+(-140)^2+(-89)^2\)
\(=1600 + 196+132496+6561+19600+7921\)
\(=168374\)
\(s=\sqrt{\frac{168374}{5}}\approx183.5\)

The test statistic for a one - sample \(t\) - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\)
Here, \(\mu = 1000\), \(\bar{x}=660\), \(s\approx183.5\), \(n = 6\)
\(t=\frac{660 - 1000}{183.5/\sqrt{6}}=\frac{-340}{74.9}\approx - 4.54\)

Answer:

B. \(H_0:\mu = 1000\) hic, \(H_1:\mu<1000\) hic