Using the t-distribution, as we have the standard deviation for the samples, it is found that since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, we test if they are equally effective teachers, that is, the subtraction of their means is 0, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, we test if they are not equally effective teachers, that is, the subtraction of their means is not 0, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

<h3>What is the distribution of the differences?</h3>

For Mrs. Elite, we have that:

$\mu_1 = 78, \sigma_1 = 10, n_1 = 30, s_1 = \frac{10}{\sqrt{30}} = 1.82574$

For Mrs. Bright, we have that:

$\mu_2 = 85, \sigma_2 = 15, n_2 = 25, s_2 = \frac{15}{\sqrt{25}} = 3$

For the distribution of differences, we have that:

$\overline{x} = \mu_1 - \mu_2 = 78 - 85 = -7$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{1.82574^2 + 3^2} = 3.5119$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{-7 - 0}{3.5119}$

$t = -1.99$

Considering a<em> two-tailed test</em>, as we are testing if the mean is different of a value, with 30 + 25 - 2 = <em>53 df and a significance level of 0.1</em>, the critical value is of $|t^{\ast}| = 1.6741$ .

Since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

To learn more about the t-distribution, you can take a look at brainly.com/question/13873630

3 0

3 years ago

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6262 - 5656}{333.3}$

$Z = 1.82$

$Z = 1.81$ has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

$P = 100% - 96.562% = 3.438%$ of the light bulbs will last more than 6262 hours.

(b) What proportion of light bulbs will last 5252 hours or less?

This is the pvalue of the zscore of $X = 5252$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5252- 5656}{333.3}$

$Z = -1.21$

$Z = -1.21$ has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 hours?

This is the pvalue of the zscore of $X = 6262$ subtracted by the pvalue of the zscore $X = 5858$

For $X = 6262$ , we have that $Z = 1.81$ with a pvalue of .96562.

For X = 5858

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5858- 5656}{333.3}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

$P = .96562 - .72907 = .23655$

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

(d) What is the probability that a randomly selected light bulb lasts less than 4646 hours?

This is the pvalue of the zscore of $X = 4646$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4646- 5656}{333.3}$

$Z = -3.03$

$Z = -3.03$ has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0

3 years ago

Please help me with question 1 (Trig Practice)

ipn [44]

Check the picture below.

8 0

3 years ago

Please help me write an equation for this polynomial graphed

oksian1 [2.3K]

The equation for this polynomial graphed will be P(x) = a(x + 2)(x + 1)(x – 3)³.

<h3>What is polynomial?</h3>

A polynomial expression is an algebraic expression with variables and coefficients.

The graph of the polynomial is given below.

From the graph, the roots of the polynomial will be -2, -1, and 3.

Then the factor of the polynomial will be

(x + 2), (x + 1), and (x – 3)

Then the polynomial will be

P(x) = a(x + 2)(x + 1)(x – 3)³

More about the polynomial link is given below.

brainly.com/question/17822016

#SPJ1

3 0

2 years ago

Read 2 more answers