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3 years ago

Which describes a statistic of a sample?

Mathematics

1 answer:

sesenic [268]3 years ago

6 0

D. Because the rest don't involve necessarily statistics and a sample would be a portion of something

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The temperature changed from 7°f at 6pm to -5°f at midnight. What was the difference between the high an low temperatures? Whats

Ronch [10]

Answer:

Difference between high and low temperatures: 12°F

Average change in temperature per hour: 2°F

Step-by-step explanation:

The difference between high and low temperatures was: 7 - (-5) = 12 (°F)

Midnight is 12am, so 6pm to midnight is 6 hours.

The average change in temperature per hour is: 12 / 6 = 2 (°F)

7 0

3 years ago

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

3 years ago

A pool in the shape of a rectangle prism is being filled with water. The length and width is 24 feet and 15 feet. If the hight o

hram777 [196]

Check the picture below.

$\textit{volume of a rectangular prism}\\\\ V=Lwh~~ \begin{cases} L=length\\ w=width\\ h=height\\[-0.5em] \hrulefill\\ L=24\\ w=15\\ h=1\frac{1}{3} \end{cases}\implies V=(24)(15)(1\frac{1}{3})\implies \stackrel{in~ft^3}{V}=(24)(15)(\frac{4}{3}) \\\\\\ \stackrel{\textit{converting all feet to inches}}{\stackrel{in~in^3}{V}=(24\cdot 12)(15\cdot 12)(\frac{4}{3}\cdot 12)}\implies V=(288)(180)(16)\implies V=\stackrel{in^3}{829440}$