Question

ssume 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 0.05 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? 760     760     1230     638     594     674

what are the null and alternative hypotheses?

determine the test statistic.

determine the P-value

state the final conclusion that addresses the original claim.

Answer 1

Solution

Final answers are given below. Back-up Theory and Details of calculations follow at the end.

Hypotheses:

Null H₀: µ = µ₀ = 1000 Vs Alternative H_A: µ < 1000 Answer 1

Test statistic: -2.3647 Answer 2

P-value: 0.0322 Answer 3

Final conclusion:

All of the child booster seats meet the specified​ requirement. Answer 4

Back-up Theory and Details of calculations

Let X = hic measurement

Let µ and σ be respectively the mean and standard deviation of X

Claim: All of the child booster seats meet the specified​ requirement.

Hypotheses:

Null H₀: µ = µ₀ = 1000 Vs Alternative H_A: µ < 1000

Test statistic:

t = (√n)(Xbar - µ₀)/s,

where

n = sample size;

Xbar = sample average;

s = sample standard deviation.

Summary of Excel Calculations is given below:

i	xi
1	760
2	760
3	1230
4	638
5	594
6	674

n	6
µ0	1000
Xbar	776.0000
s	232.0345
tcal	-2.3647
Given α	0.05
tcrit	-2.0150
p-value	0.032188

Distribution, significance level, α , Critical Value and p-value

Under H₀, t ~ t_{n – 1}

Critical value = lower α% point of t_{n – 1}.

p-value = P(t_{n – 1} < tcal)

Using Excel Function, Statistical TDIST p-value is found to be as shown in the above table.

Decision:

Since tcal > tcrit, or equivalently since p-value < α. H₀ is rejected.

Conclusion:

There is sufficient evidence to support the claim.

DONE