Someone please help!!

Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab

Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected, and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, $H_0$ : $\mu$ = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, $H_1$ : $\mu$ < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

T.S. = $\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, X bar = sample mean = 738.44 hours

s = sample standard deviation = 38.2 hours

n = sample size = 50

So, test statistics = $\frac{738.44-750}{\frac{38.2}{\sqrt{50} } }$ ~ $t_4_9$

= -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0

3 years ago

Transpose v= u + t for t

LiRa [457]

Uv al dt or trace mcsorely

4 0

3 years ago

(2+x)dy/dx=3y need help

den301095 [7]

$(2+x) \frac{dy}{dx} =3y$
$\frac{1}{3y} dy= \frac{1}{2+x} dx$
$\int { \frac{1}{3y} } \, dy = \int { \frac{1}{2+x} } \, dx$
$\frac{1}{3}\ln(3y) = \ln(2+x) + C$
(raise both to the power e): $e^{ \frac{1}{3}\ln(3y)}=e^{\ln(k(2+x))}$
$(3y)^ \frac{1}{3} =k(2+x)$
$3y=k(2+x)^3$
$y= k(2+x)^3$

5 0

3 years ago

A triangle has side lengths of 33,56, and 65. Is it a right triangle? Explain.

soldier1979 [14.2K]

U can see if it is a right triangle bu using the pythagorean theorem (or however it is spelled)

a^2 + b^2 = c^2....where a and b are the legs and c is the hypotenuse (the largest side)

so lets sub our numbers in and see if they equal
33^2 + 56^2 = 65^2
1089 + 3136 = 4225
4225 = 4225.......correct...so YES, this is a right triangle

5 0

2 years ago

Read 2 more answers

Select all of the points that are solutions to the system of linear inequalities that is listed below 10x + 4y < 12 8x -3y &g

liberstina [14]

Answer:

(3, -8) and (2, -10)

Step-by-step explanation:

Given

$10x + 4y < 12$