3. The producers of a new toothpaste want to see if it is performing similar to their main whiteningtoothpaste in preventing cavities. Their main whitening tooth paste performance at a six-monthcheckup is i.73 (o-1.12). To test the new tooth paste, 60 people used it for six-months. At thecheckup they found a mean of 1.5 cavities. Perform the appropriate hypothesis testing. Makesure to include all the steps (hypotheses, critical values, calculations, conclusions aninterpretations).0.068

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Asked Oct 4, 2019
3. The producers of a new toothpaste want to see if it is performing similar to their main whitening
toothpaste in preventing cavities. Their main whitening tooth paste performance at a six-month
checkup is i.73 (o-1.12). To test the new tooth paste, 60 people used it for six-months. At the
checkup they found a mean of 1.5 cavities. Perform the appropriate hypothesis testing. Make
sure to include all the steps (hypotheses, critical values, calculations, conclusions an
interpretations).
0.0
68
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3. The producers of a new toothpaste want to see if it is performing similar to their main whitening toothpaste in preventing cavities. Their main whitening tooth paste performance at a six-month checkup is i.73 (o-1.12). To test the new tooth paste, 60 people used it for six-months. At the checkup they found a mean of 1.5 cavities. Perform the appropriate hypothesis testing. Make sure to include all the steps (hypotheses, critical values, calculations, conclusions an interpretations). 0.0 68

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Step 1

Hypotheses:

Null hypothesis:

H0: µ = 1.73

That is, the new toothpaste performs similar to main whitening toothpaste in preventing cavities.

 

Alternative hypothesis:

H1: µ ≠1.73

That is, the new toothpaste performs is not similar to main whitening toothpaste in preventing cavities.

Since the alternative hypothesis states µ ≠1.73, this test is a two-tailed test.

Step 2

Critical value:

Consider the level of significance as 0.05.The critical value for 0.05 significance level is 1.96.

Step 3

Calculation:

It is given that X has a normal distribution and σ = 1.12.

Since it is assumed that X has a normal distribution with known σ, the standard normal is the appropriate sampling distribution.

Test statistic for z-test:

Here, the sample mean, x-bar is 1.5.

Population mean, µ is 1.73.

...

Z
1.5 1.73
1.12
60
-0.23
0.1446
=-1.59
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Z 1.5 1.73 1.12 60 -0.23 0.1446 =-1.59

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