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AVprozaik [17]
3 years ago
10

The Walt Disney Company, the production studio behind the Marvel Cinematic Universe, is trying to find the best release date for

their new super hero movie. They are hesitating between the summer and winter holiday release, and they want to know if there is any difference between the earning potential during those seasons. The analytics division in the company examined a random sample of box-office data for movies released during the past 5 years in the USA and Canada and found that 52 out of 1258 movies with summer release date earned over 400 million dollars. They also counted that out of 545 movies released in the winter, 8 earned over 400 million dollars.
Over 4 million USD Under 4 million USD
Summer release 57 1251
Winter release 11 544
Total
1. We want to investigate whether there is a difference in the proportion of movies that eam over 400 million dollars for the two release seasons. Which hypotheses should we use?
2. Calculate the difference in the proportions of movies that eamed over 400 million dollars.
3. The paragraph below describes the set up for a randomization test, if we were to conduct a hypothesis test without using software. Fill in the blanks with a number.
We write Summer on_____cards and cards representing the movies with summer release date, and Winter on_____cards. Then, we shuffle these cards and split them into two groups one group of size_____representing the movies with box-office over 400 million dollars, and another group of size_____representing the rest of the movies. We calculate the difference in the proportions of movies that eaned over 400 million dollars during the two release seasons to get PSmmer PWinter. Finally, we build a histogram of these simulated dfterences.
Mathematics
1 answer:
melomori [17]3 years ago
7 0

Answer:

Step-by-step explanation:

                           Over 4 million USD    Under 4 million USD    Total

Summer release      57                               1195                         1251

Winter release         11                                533                           544

1)Let P_1 be the proportion of movies that earn over 400 million in summer and P_2 be the proportion of movies that earn over 400 million in winter.

Null hypothesis:H_0:P_1= P_2

Alternate hypothesis : H_a: P_1 \neq P_2

2)Calculate the difference in the proportions of movies that earned over 400 million dollars.

P_1=\frac{57}{1251}=0.045

P_2=\frac{11}{544}=0.0202

3)

P_{(Summer)}-P_{(winter)}=0.045-0.0202=0.0248

We write Summer on 1251 cards and cards representing the movies with summer release date, and Winter on 544 cards. Then, we shuffle these cards and split them into two groups one group of size 68 representing the movies with box-office over 400 million dollars, and another group of size representing 1728 the rest of the movies.

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The feet of the average adult woman are 24.6 cm long, and foot lengths are normally distributed. If 16% of adult women have feet
nevsk [136]

Answer:

Approximately 16% of adult women have feet longer than 27.2 cm.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The feet of the average adult woman are 24.6 cm long

This means that \mu = 24.6

16% of adult women have feet that are shorter than 22 cm

This means that when X = 22, Z has a p-value of 0.16, so when X = 22, Z = -1. We use this to find \sigma

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

-1 = \frac{22 - 24.6}{\sigma}

-\sigma = -2.6

\sigma = 2.6

Approximately what percent of adult women have feet longer than 27.2 cm?

The proportion is 1 subtracted by the p-value of Z when X = 27.2. So

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

Z = \frac{27.2 - 24.6}{2.6}

Z = 1

Z = 1 has a p-value of 0.84.

1 - 0.84 = 0.16

0.16*100% = 16%.

Approximately 16% of adult women have feet longer than 27.2 cm.

3 0
3 years ago
Can y’all help pls ;)
Vikentia [17]

Answer:

A. 5x+6

Step-by-step explanation:

-\frac{1}{3}(9x-18)+8x

-3x+6+8x (multiply across the -1/3)

5x+6 (combine 8x and -3x)

4 0
3 years ago
A newspaper reports the median house price is $250,000. From that standard, tell what you know about housing prices.
Alik [6]
You know that the average house price is $250,000 dollars
Median means medium
8 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 F and a standard deviation of 0.56 F. Co
Ira Lisetskai [31]

Answer:

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

Step-by-step explanation:

The first step is finding the confidence interval

The sample size is 103.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

df = 103-1 = 102

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have T = 2.63.

Now, we need to find the standard deviation of the sample. That is:

s = \frac{0.56}{\sqrt{103}} = 0.055

Now, we multiply T and s

M = T*s = 2.63*0.055 = 0.145

For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.

For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

5 0
3 years ago
15÷276 i need to know for math homework its kinda hard to do​
Stells [14]

Answer:

0.0543

(i am not fully sure)

5 0
2 years ago
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