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kondaur [170]
4 years ago
10

For the next three questions use the following information to determine your answers. A research group is curious about features

that can be attributed to music genres. A music streaming service provides a few different attributes for songs such as speechness, danceability, and valence. They suspect that there is a difference between the average valence (positive or negative emotion) of metal songs compared to blues songs. However, they must conduct a study to determine if that is true. From a sample of 87 metal songs, the sample mean for valence is 0.451 and the sample standard deviation is 0.139. From a sample of 94 blues songs, the sample mean for valence is 0.581 and the sample standard deviation is 0.167. Assume that sample1 comes from the sample metal songs and that sample2 comes from the sample blues songs Compute the 90% confidence interval. Please round the values to the fourth decimal point and format your response as follows: (lower_value, upper_value)
Which of the following represents the hypotheses that we will be testing, assuming that µ1 represents the population mean of valence for all metal songs and that µ2 represents the population mean of valence for all blues songs.?

a.H0: µ1 = µ2 versus Ha: µ1 > µ2

b.H0: µ1 = µ2 versus Ha: µ1 ≠ µ2

c.H0: µ1 = µ2 versus Ha: µ1 < µ2
Mathematics
2 answers:
ASHA 777 [7]4 years ago
7 0

Answer:

b

Step-by-step explanation:

Null hypothesis: mean of valence for all metal song is equal to mean of valence of blues song.

Null hypthesis tests the claim

Altenate hypothesis: mean of valence for all metal songs is not equal to mean of valence of blues song.

Alternate hypothesis rejects the claim.

rewona [7]4 years ago
6 0

Answer:

(A) Since Sample 1 comes from the population of Metal songs only and Sample 2 comes from the population of Blue songs only,

The 90% confidence interval for mean valence for Metal songs is (0.2224 , 0.6797)

The 90% confidence interval for mean valence for Blue songs is (0.3063 , 0.8557)

All intermittent and final answers were rounded up to four decimal places.

(B) The option (b) is correct. This is because the question says that the research group suspects a difference between both means, not that one mean is greater or less than the other.

Step-by-step explanation:

(A) Using a 90% confidence level, the true mean is within 1.645 standard deviations of the sample mean

For METAL SONGS,

Lower limit = 0.451 - (1.645)(0.139)

= 0.451 - 0.2287 = 0.2224

Upper limit = 0.451 + (1.645)(0.139)

= 0.451 + 0.2287 = 0.6797

For BLUE SONGS,

Lower limit = 0.581 - (1.645)(0.167)

= 0.581 - 0.2747 = 0.3063

Upper limit = 0.581 + (1.645)(0.167)

= 0.581 + 0.2747 = 0.8557

(B) The null hypothesis is:

Mean valence of Metal songs is equal to the mean valence of Blue songs

Alternative hypothesis:

Mean valence of Metal songs is NOT equal to the mean valence of Blue songs.

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Consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs show the p
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Answer:

Regression function: y=1986.406+0.0059x

The function predicts that population will reach 14,000 in year 2068.

Step-by-step explanation:

We have to determine a function y=b_0+b_1x_1 by applying linear regression. The data we have is 5 pair of points which relates population to year.

According to the simple regression model (one independent variable), if we minimize the error between the model (the linear function) and the points given, the parameters are:

b_0=\bar{y}+b_1\bar{x}\\\\b_1=\frac{\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}}{\sum\limits^5_{i=1} {(x_i-\bar x)^2}}

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\bar x=\frac{2500+2650+3000+3500+4200}{5}=\frac{15850}{5}=3170\\\\ \bar y=\frac{2001+2002+2004+2007+2011}{5}=\frac{10025}{5}=2005

The sample covariance can be calculated as

\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}=(2500-3170)(2001-2005)+(2650-3170)(2002-2005)+(3000-3170)(2004-2005)+(3500-3170)(2007-2005)+(4200-3170)(2011-2005)\\\\\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}=2680+1560+170+660+6180\\\\ \sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}=11250

The variance of x can be calculated as

\sum\limits^5_{i=1} {(x_i-\bar x)^2}=(2500-3170)^2+(2650-3170)^2+(3000-3170)^2+(3500-3170)^2+(4200-3170)^2\\\\\sum\limits^5_{i=1} {(x_i-\bar x)^2}=448900+270400+28900+108900+1060900\\\\\sum\limits^5_{i=1} {(x_i-\bar x)^2}=1918000

Now we can calculate the parameters of the regression model

b_1=\frac{\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}}{\sum\limits^5_{i=1} {(x_i-\bar x)^2}}=\frac{11250}{1918000}=0.005865485  \\\\ b_0=\bar{y}+b_1\bar{x}=2005-0.005865485*3170=1986.406413

The function then become:

y=1986.406+0.0059x

With this linear equation we can predict when the population will reach 14,000:

y=1986.406+0.0059(14,000)=1986.406+82.117=2068.523

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2x^2+7x-4

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