Given: 5x+6−x=2x+10

A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute

GREYUIT [131]

Answer:

(a) Null Hypothesis, $H_0$ : $\sigma$ = 10 beats per minute

Alternate Hypothesis, $H_A$ : $\sigma\neq$ 10 beats per minute

(b) The value of chi-square test statistics is 35.704.

(c) P-value = 0.4360.

(d) We conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

Step-by-step explanation:

We are given that a simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute.

If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute.

Let $\sigma$ = population standard deviation for the pulse rates of men.

(a) So, Null Hypothesis, $H_0$ : $\sigma$ = 10 beats per minute {means that the pulse rates of men have a standard deviation equal to 10 beats per minute}

Alternate Hypothesis, $H_A$ : $\sigma\neq$ 10 beats per minute {means that the pulse rates of men have a standard deviation different from 10 beats per minute}

The test statistics that will be used here is One-sample chi-square test for standard deviation;

T.S. = $\frac{(n-1)\times s^{2} }{\sigma^{2} }$ ~ $\chi^{2}__n_-_1$

where, s = sample standard deviation = 10.1 beats per minute

n = sample of men = 36

So, the test statistics = $\frac{(36-1)\times 10.1^{2} }{10^{2} }$ ~ $\chi^{2}__3_5$

= 35.70 4

(b) The value of chi-square test statistics is 35.704.

(c) Also, the P-value of the test statistics is given by;

P-value = P( $\chi^{2}__3_5$ > 35.704) = 0.4360

(d) Since the P-value of our test statistics is more than the level of significance as 0.4360 > 0.10, so we have insufficient evidence to reject our null hypothesis as the test statistics will not fall in the rejection region.

Therefore, we conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

3 0

4 years ago

Factoring can only be done with trinomials. Group of answer choices True False

elena-s [515]

Answer:

True

Step-by-step explanation:

It’s true

4 0

3 years ago

How can I express logbx-(logby+logbz) as a single log?

GuDViN [60]

$\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)
\\ \quad \\
% Logarithm of rationals
log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)\\\\
-----------------------------\\\\
log_b(x)-[log_b(y)+log_b(z)]\implies log_b(x)-[log_b(yz)]
\\\\\\
log_b\left( \cfrac{x}{yz} \right)$

3 0

4 years ago

What is the value of k in the equation? 20^k/ 20^4 = 20^9

Luden [163]

Answer:

k = 13

Step-by-step explanation:

20 ^ k / 20^4 = 20^9

knowing the laws of exponents, diving two numbers of the same base with exponents is like subtracting the exponents, so

k^x / k^y

= k ^ (x - y)

apply that knowledge here

20 ^ k / 20 ^ 4 = 20 ^ 9

20 ^ ( k - 4 ) = 20 ^ 9

now the bases have no use here so ignore them, and just calculate the exponents

k - 4 = 9

solve using inverse operations

k - 4 = 9

+4 +4

k = 13

4 0

3 years ago

Read 2 more answers

Rewrite 3x3x3x3x3 using an exponent

cluponka [151]

Answer:

3^5

Step-by-step explanation:

Every time a number is multiplied by itself, that adds another power

3 = 3^1

3*3 = 3^2

3*3*3 = 3^3

PROBLEM: 3*3*3*3*3 = 3^5

Answer: 3^5

5 0

3 years ago

Read 2 more answers