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nikitadnepr [17]

2 years ago

5

Josie is training for a race. Pthe radio of the number of minutes she runs to the number of miles she runs is 24 to 3. she plans

to run 10 miles how many minutes will it take her?

1 answer:

tatyana61 [14]2 years ago

3 0

It will take 2 hours

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Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? The ar

snow_tiger [21]

Answer:

Yes. There is enough evidence to support the claim that the remaining toothpaste is significantly is less than 10% of the advertised net content.

Step-by-step explanation:

The sample of the remaining toothpaste is: [.53, .65, .46, .50, .37].

This sample has a size n=5, a mean M=0.502 and standard deviation s=0.102.

$M=\dfrac{1}{5}\sum_{i=1}^{5}(0.53+0.65+0.46+0.5+0.37)\\\\\\ M=\dfrac{2.51}{5}=0.502$

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{5}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}\cdot [(0.53-(0.502))^2+...+(0.37-(0.502))^2]}\\\\\\s=\sqrt{\dfrac{1}{4}\cdot [(0.001)+(0.022)+(0.002)+(0)+(0.017)]}\\\\\\ s=\sqrt{\dfrac{0.04188}{4}}=\sqrt{0.01047}\\\\\\s=0.102$

The 10% of the advertised content is:

$0.10\cdot 6.0\;oz=0.6\:oz$

Hypothesis test for the population mean:

The claim is that the remaining toothpaste is significantly is less than 10% of the advertised net content.

Then, the null and alternative hypothesis are:

$H_0: \mu=0.6\\\\H_a:\mu< 0.6$

The significance level is 0.05.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.102}{\sqrt{5}}=0.046$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{0.502-0.6}{0.046}=\dfrac{-0.098}{0.046}=-2.148$

The degrees of freedom for this sample size are:

$df=n-1=5-1=4$

This test is a left-tailed test, with 4 degrees of freedom and t=-2.148, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t$

As the P-value (0.049) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the remaining toothpaste is significantly is less than 10% of the advertised net content.

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