An employment information service claims the mean annual salary for senior level product engineers is $96,000. The annual salaries (in dollars) for a random sample of 16 senior level product engineers are shown in the table to the right. At α=0.01, test the claim that the mean salary is $96,000. Complete parts (a) through (e) below. Assume the population is normally distributed. Annual Salaries 100 ,703 96 ,223 93 ,487 112 ,642 82 ,478 74 ,184 76 ,998 81 ,040 102 ,503 76 ,223 104 ,123 103 ,943 91 ,025 82 ,062 85 ,032 110 ,249 (a) Identify the claim and state H0 and Ha. H0: ▼ sigma squaredσ2 muμ sigmaσ pp ▼ greater than> greater than or equals≥ equals= not equals≠ less than< less than or equals≤ nothing Ha: ▼ sigma squaredσ2 muμ sigmaσ pp ▼ equals= greater than> not equals≠ less than< less than or equals≤ greater than or equals≥ nothing (Type integers or decimals. Do not round.) The claim is the ▼ alternative null hypothesis. (b) Use technology to find the critical value(s) and identify the rejection region(s). The critical value(s) is/are t0=nothing. (Use a comma to separate answers as needed. Round to two decimal places as needed.) (c) Find the standardized test statistic, t. The standardized test statistic is t=nothing. (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. ▼ Fail to reject Reject H0 because the standardized test statistic ▼ is is not in the rejection region. (e) Interpret the decision in the context of the original claim. There ▼ is not is enough evidence at the nothing% level of significance to ▼ rejectreject supportsupport the claim that the mean annual salary for senior level product engineers is ▼ greater than equal to less than not equal to less than or equal to greater than or equal to $nothing. (Type integers or decimals. Do not round.)

An employment information service claims the mean annual salary for senior level product engineers is $96,000. The annual salaries (in dollars) for a random sample of 16 senior level product engineers are shown in the table to the right. At α=0.01, test the claim that the mean salary is $96,000. Complete parts (a) through (e) below. Assume the population is normally distributed. Annual Salaries 100 ,703 96 ,223 93 ,487 112 ,642 82 ,478 74 ,184 76 ,998 81 ,040 102 ,503 76 ,223 104 ,123 103 ,943 91 ,025 82 ,062 85 ,032 110 ,249 (a) Identify the claim and state H0 and Ha. H0: ▼ sigma squaredσ2 muμ sigmaσ pp ▼ greater than> greater than or equals≥ equals= not equals≠ less than< less than or equals≤ nothing Ha: ▼ sigma squaredσ2 muμ sigmaσ pp ▼ equals= greater than> not equals≠ less than< less than or equals≤ greater than or equals≥ nothing (Type integers or decimals. Do not round.) The claim is the ▼ alternative null hypothesis. (b) Use technology to find the critical value(s) and identify the rejection region(s). The critical value(s) is/are t0=nothing. (Use a comma to separate answers as needed. Round to two decimal places as needed.) (c) Find the standardized test statistic, t. The standardized test statistic is t=nothing. (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. ▼ Fail to reject Reject H0 because the standardized test statistic ▼ is is not in the rejection region. (e) Interpret the decision in the context of the original claim. There ▼ is not is enough evidence at the nothing% level of significance to ▼ rejectreject supportsupport the claim that the mean annual salary for senior level product engineers is ▼ greater than equal to less than not equal to less than or equal to greater than or equal to $nothing. (Type integers or decimals. Do not round.)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

An employment information service claims the mean annual salary for senior level product engineers is $96,000. The annual salaries (in dollars) for a random sample of 16 senior level product engineers are shown in the table to the right. At α=0.01, test the claim that the mean salary is $96,000. Complete parts (a) through (e) below. Assume the population is normally distributed.
Annual Salaries
100	,703	96	,223	93	,487	112	,642
82	,478	74	,184	76	,998	81	,040
102	,503	76	,223	104	,123	103	,943
91	,025	82	,062	85	,032	110	,249

(a) Identify the claim and state

and

Ha.

H0:	▼ sigma squaredσ2 muμ sigmaσ pp ▼ greater than> greater than or equals≥ equals= not equals≠ less than< less than or equals≤ nothing
Ha:	▼ sigma squaredσ2 muμ sigmaσ pp ▼ equals= greater than> not equals≠ less than< less than or equals≤ greater than or equals≥ nothing

(Type integers or decimals. Do not round.)

The claim is the

▼

alternative

null

hypothesis.

(b) Use technology to find the critical value(s) and identify the rejection region(s).

The critical value(s) is/are

t0=nothing.

(Use a comma to separate answers as needed. Round to two decimal places as needed.)

The standardized test statistic is

t=nothing.

(Round to two decimal places as needed.)

(d) Decide whether to reject or fail to reject the null hypothesis.

▼

Fail to reject

Reject

because the standardized test statistic

▼

is not

in the rejection region.

(e) Interpret the decision in the context of the original claim.

There

▼

is not

enough evidence at the

nothing%

level of significance to

▼

rejectreject

supportsupport

the claim that the mean annual salary for senior level product engineers is

▼

greater than

equal to

less than

not equal to

less than or equal to

greater than or equal to

$nothing.

(Type integers or decimals. Do not round.)

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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