I am going 60 miles away and I am driving 40 miles per hour the whole way how long will it take me to get there

A student carried out an experiment to determine the amount of vitamin C in a tablet sample. He performed 5 trials to produce th

ivolga24 [154]

Answer:

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

P-value = 0.166.

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{5}(490+502+505+495+492)\\\\\\M=\dfrac{2484}{5}\\\\\\M=496.8\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}((490-496.8)^2+(502-496.8)^2+(505-496.8)^2+(495-496.8)^2+(492-496.8)^2)}\\\\\\s=\sqrt{\dfrac{166.8}{4}}\\\\\\s=\sqrt{41.7}=6.5\\\\\\$

Then, we can perform the hypothesis t-test for the mean.

The claim is that the amount of vitamin C in a tablet sample is different from 500 mg.

Then, the null and alternative hypothesis are:

$H_0: \mu=500\\\\H_a:\mu< 500$

The significance level is 0.05.

The sample has a size n=5.

The sample mean is M=496.8.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.5.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6.5}{\sqrt{5}}=2.907$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{496.8-500}{2.907}=\dfrac{-3.2}{2.907}=-1.1$

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=-1.1, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.166) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

4 0

2 years ago

Mr.Eastman drew triangle JKL,with a height of 12 inches and a base of 12 inches, and triangle XLZ, with a height of 8 inches and

34kurt

Answer:

Step-by-step explanation:

Area of triangle = 1/2 × Base × Height

Mr.Eastman drew triangle JKL,with a height of 12 inches and a base of 12 inches

Area of JKL = 1/2 × 12 inches × 12 inches

= 72 Square inches

triangle XLZ, with a height of 8 inches and a base of 16 inches. Which triangle has the greater area?

5 0

3 years ago

Which table goes with the graph?

Fudgin [204]

Answer:

74

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Soft drink consumption in New Zealand. A survey commissioned by the Southern Cross Healthcare Group reported that 16% of New Zea

lidiya [134]

<h2>Answer with explanation:</h2>

Given : The proportion of New Zealanders consume five or more servings of soft drinks per week : $\hat{p}= 0.16$

a) The number of survey respondents reported that they consume five or more servings of soft drinks per week = 2006

b) Confidence interval for population proportion (p) :

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$ = Sample proportion.

n= Sample size.

z* = Critical value.

For n= 2006 , $\hat{p}= 0.16$ and critical value for 95% confidence interval : z* = 1.96

Then , the required confidence interval will be :

$0.16\pm (1.96)\sqrt{\dfrac{ 0.16(1- 0.16)}{2006}}$

$0.16\pm (1.96)\sqrt{ 0.000066999002991}$

$0.16\pm (1.96)(0.00818529186963)$

$0.16\pm 0.0160431720645$

$(0.16-0.0160431720645,\ 0.16+0.0160431720645)$

$(0.143956827935,\ 0.176043172064)\approx(0.144,\ 0.176)$

i.e. A 95% confidence interval for the proportion of New Zealanders who report that they consume five or more servings of soft drinks per week. = $(0.144,\ 0.176)$

c) $(0.144,\ 0.176)=(0.144\times100\%,\ 0.176\times100\%)$

$=(14.4\%,\ 17.6\%)$ 176\times100\%)[/tex]

d) The estimate might be biased because the survey is taken online that means offline New Zealanders are out of consideration.

6 0

3 years ago

Use the place-valué charro to quite a Numbers in the table that is 1/10 of given Number?

Lana71 [14]

This is the answer for the Use the place-valué charro to quite a Numbers in the table that is 1/10 of given Number?

6 0

3 years ago