Answer is center of rotation

"An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particul

Galina-37 [17]

Answer:

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15) = 0.0173

b) Probability of not rejecting the claim when p = 0.7, P(X > 15) = 0.8106

when p = 0.6, P(X > 15) = 0.4246

c) Check Explanation

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Step-by-step explanation:

p is the true proportion of houses with smoke detectors and p = 0.80

The claim that 80% of houses have smoke detectors is rejected if in a sample of 25 houses, not more than 15 houses have smoke detectors.

If X is the number of homes with detectors among the 25 sampled

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15)

This is a binomial distribution problem

A binomial experiment is one in which the probability of success doesn't change with every run or number of trials (probability that each house has a detector is 0.80)

It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure (we are sampling 25 houses with each of them either having or not having a detector)

The outcome of each trial/run of a binomial experiment is independent of one another.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = less than or equal to 15

p = probability of success = probability that a house has smoke detectors = 0.80

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.80 = 0.20

P(X ≤ 15) = Sum of probabilities from P(X = 0) to P(X = 15) = 0.01733186954 = 0.01733

b) Probability of not rejecting the claim when p= 0.7 when p= 0.6

For us not to reject the claim, we need more than 15 houses with detectors, hence, th is probability = P(X > 15), but p = 0.7 and 0.6 respectively for this question.

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = more than 15

p = probability that a house has smoke detectors = 0.70, then 0.60

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.70 = 0.30

And 1 - 0.60 = 0.40

P(X > 15) = sum of probabilities from P(X = 15) to P(X = 25)

When p = 0.70, P(X > 15) = 0.8105639765 = 0.8106

When p = 0.60, P(X > 15) = 0.42461701767 = 0.4246

c) How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14.

The error probabilities include the probability of the claim being false.

When X = 15

(Error probability when p = 0.80) = 0.0173

when p = 0.70, error probability = P(X ≤ 15) = 1 - P(X > 15) = 1 - 0.8106 = 0.1894

when p = 0.60, error probability = 1 - 0.4246 = 0.5754

When X = 14

(Error probability when p = 0.80) = P(X ≤ 14) = 0.00555

when p = 0.70, error probability = P(X ≤ 14) = 0.0978

when p = 0.60, error probability = P(X ≤ 14) = 0.4142

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Hope this Helps!!!

6 0

2 years ago

A recent study found that 51 children who watched a commercial for Walker Crisps (potato chips) featuring a long-standing sports

hichkok12 [17]

Answer:

1

The claim is that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

2

The kind of test to use is a t -test because a t -test is used to check if there is a difference between means of a population

3

$t = 3.054$

4

The p-value is $p-value = P(Z > 3.054) = 0.0011291$

5

The conclusion is

There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

The test statistics is

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 51$

The first sample mean is $\mu_1 = 36$

The second sample size is $n_2 = 41$

The second sample size is $\mu_2 = 25$

The first standard deviation is $\sigma _1 = 21.4 \ g$

The second standard deviation is $\sigma _2 = 12.8 \ g$

The level of significance is $\alpha = 0.05$

The null hypothesis is $H_o : \mu_1 = \mu_ 2$

The alternative hypothesis is $H_a : \mu_1 > \mu_2$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } }$

=> $t = \frac{ 36 - 25}{ \sqrt{ \frac{ 21.4^2}{51} + \frac{ 12.8^2}{41} } }$

=> $t = 3.054$

The p-value is mathematically represented as

$p-value = P(Z > 3.054)$

Generally from the z table

$P(Z > 3.054) = 0.0011291$

=> $p-value = P(Z > 3.054) = 0.0011291$

From the values obtained we see that $p-value < \alpha$ so the null hypothesis is rejected

Thus the conclusion is

There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

5 0

2 years ago

The Andersons have a booth at the farmers market they set up each weekend over the summer they sell homemade apple pie for $9.75

butalik [34]

Answer:

The money they make on selling pies on Saturday is $68 (approximately).

Step-by-step explanation:

Given:

Cost of an apple pie is $9.75.

Total apple pie sold on Saturday is 7.

So, to get the total amount of apple pie sold on Saturday:

$Cost\ of\ an\ apple\ pie\times Number\ of\ apple\ pie\ sold$