The first pic is the question. The second pic is an example of how to solve the problem so that you can better answer it correctly. Cost of Braces The average cost for teeth straightening with metal bracesis approximately s5350. A nationwide franchise thinks that its cost is belowthat figure. A random sample of 27 patients across the country had anaverage cost of s5153 with a standard deviation of s631. At a=0.10, can it beconcluded that the mean is less than s5350? Assume that the population isapproximately normally distributed.Part: 0 / 5Part 1 of 5(a) State the hypotheses and identify the claim.Ho: H (Choose one) ▼$(Choose one) ▼H: µ (Choose one) ▼$(Choose one) ▼This hypothesis test is a (Choose one) ▼ test. Sample QuestionCost of Braces The average cost for teeth straighteningwith metal braces is approximately ss406. A nationwidefranchise thinks that its cost is below that figure. A randomsample of 26 patients across the country had an averagecost of ss230 with a standard deviation of $629. At a-0.005, canit be concluded that the mean is less than ss406? Assumethat the population is approximately normally distributed.(a) State the hypotheses and identify the daim.(b) Find the critical value.(c) Compute the test value.(d) Make the decision.(e) Summarize the results.Explanation(a) State the hypotheses and identify the claim.The null hypothesis H, is the statement that there is nodifference between a parameter and a specific value. This isequivalent to H:- 5406 dollars.The alternative hypothesis H, is the statement that there isa difference between a parameter and a specific value. In thiscase, the mean cost for teeth straightening with metalbraces is less than $5406. This is equivalent to H, :H< 5406 dollars.The problem asks, "Is the mean cost for teeth straighteningwith metal braces less than ss406?" Hence, the claim is thealternative hypothesis.(b) Find the critical value.From OThe t Distribution Table, for a left-tailed test with a-0.005 andd.f. - 26-1-25, the critical value is --2.787.The t Distribution TableConfidenceintervals80%90%95%98%99%One tail, a0.100.050.0250.010.005d.f.Two tails, a0.200.100.050.020.01241.3181.711 2.0642.4922.797251.3161.7082.0602.4852.787261.3151.7062.0562.4792.779(c) Compute the test value.The i test is a statistical test for the mean of apopulation and is used when the population isnormally or approximately normally distributed ando is unknown.The formula for the test isThe degrees of freedom are d.f. =n-1.Using the formula for the r-test,5230 - 5406629//26--1.427Hence, the test value, rounded to 3 decimal places, isI--1.427.(d) Make the decision.Since the test value does not fall in the critical region, donot reject the null hypothesis.

The random variable cost of teeth straightening follows normal distribution. We have to test whether…

Cost of Braces The average cost for teeth straightening with metal braces is approximately s5350. A nationwide franchise thinks that its cost is below that figure. A random sample of 27 patients across the country had an average cost of s5153 with a standard deviation of s631. At a=0.10, can it be concluded that the mean is less than s5350? Assume that the population is approximately normally distributed. Part: 0 / 5 Part 1 of 5 (a) State the hypotheses and identify the claim. Ho: H (Choose one) ▼ |(Choose one) ▼ H: µ (Choose one) ▼ (Choose one) ▼ This hypothesis test is a (Choose one) ▼ test.

Cost of Braces The average cost for teeth straightening with metal braces is approximately s5350. A nationwide franchise thinks that its cost is below that figure. A random sample of 27 patients across the country had an average cost of s5153 with a standard deviation of s631. At a=0.10, can it be concluded that the mean is less than s5350? Assume that the population is approximately normally distributed. Part: 0 / 5 Part 1 of 5 (a) State the hypotheses and identify the claim. Ho: H (Choose one) ▼ |(Choose one) ▼ H: µ (Choose one) ▼ (Choose one) ▼ This hypothesis test is a (Choose one) ▼ test.

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

The first pic is the question. The second pic is an example of how to solve the problem so that you can better answer it correctly.

Cost of Braces The average cost for teeth straightening with metal braces
is approximately s5350. A nationwide franchise thinks that its cost is below
that figure. A random sample of 27 patients across the country had an
average cost of s5153 with a standard deviation of s631. At a=0.10, can it be
concluded that the mean is less than s5350? Assume that the population is
approximately normally distributed.
Part: 0 / 5
Part 1 of 5
(a) State the hypotheses and identify the claim.
Ho: H (Choose one) ▼
$
(Choose one) ▼
H: µ (Choose one) ▼
$
(Choose one) ▼
This hypothesis test is a (Choose one) ▼ test.

Sample Question
Cost of Braces The average cost for teeth straightening
with metal braces is approximately ss406. A nationwide
franchise thinks that its cost is below that figure. A random
sample of 26 patients across the country had an average
cost of ss230 with a standard deviation of $629. At a-0.005, can
it be concluded that the mean is less than ss406? Assume
that the population is approximately normally distributed.
(a) State the hypotheses and identify the daim.
(b) Find the critical value.
(c) Compute the test value.
(d) Make the decision.
(e) Summarize the results.
Explanation
(a) State the hypotheses and identify the claim.
The null hypothesis H, is the statement that there is no
difference between a parameter and a specific value. This is
equivalent to H:- 5406 dollars.
The alternative hypothesis H, is the statement that there is
a difference between a parameter and a specific value. In this
case, the mean cost for teeth straightening with metal
braces is less than $5406. This is equivalent to H, :H< 5406 dollars.
The problem asks, "Is the mean cost for teeth straightening
with metal braces less than ss406?" Hence, the claim is the
alternative hypothesis.
(b) Find the critical value.
From OThe t Distribution Table, for a left-tailed test with a-0.005 and
d.f. - 26-1-25, the critical value is --2.787.
The t Distribution Table
Confidence
intervals
80%
90%
95%
98%
99%
One tail, a
0.10
0.05
0.025
0.01
0.005
d.f.
Two tails, a
0.20
0.10
0.05
0.02
0.01
24
1.318
1.711 2.064
2.492
2.797
25
1.316
1.708
2.060
2.485
2.787
26
1.315
1.706
2.056
2.479
2.779
(c) Compute the test value.
The i test is a statistical test for the mean of a
population and is used when the population is
normally or approximately normally distributed and
o is unknown.
The formula for the test is
The degrees of freedom are d.f. =n-1.
Using the formula for the r-test,
5230 - 5406
629//26
--1.427
Hence, the test value, rounded to 3 decimal places, is
I--1.427.
(d) Make the decision.
Since the test value does not fall in the critical region, do
not reject the null hypothesis.

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