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katovenus [111]
4 years ago
7

A dieter believes that the average number of calories in a homemade peanut better cookie is more than the average number of calo

ries in a store-bought peanut better cookie. which excel statements will estimate the difference in mean calories between the two types of cookies using a 90% confidence interval. the data are: homemade store-bought sample size 40 45 mean number of calories 180 179 standard deviation 2 4
a. =(180-179)+(1.96)(0.6749), and =(180-179)-(1.96)(0.6749)
b. =(180-179)+(1.645)(0.6749), and =(180-179)-(1.645)(0.6749)
c. =(45-40)+(1.96)(0.6749), and =(45-40)-(1.96)(0.6749)
d. =(45-40)+(1.645)(0.6749), and =(45-40)-(1.645)(0.6749)
e. none of the above.
Mathematics
1 answer:
zalisa [80]4 years ago
8 0

The confidence interval formula is computed by:

Xbar ± Z s/ sqrt (n)

Where:

Xbar is the mean

Z is the z value

S is the standard deviation

N is the number of samples

 

So our given are:

90% confidence interval with a z value of 1.645

Sample size 40, 45

Mean 180, 179

Standard deviation 2, 4

 

So plugging that information in the data will give us a confidence interval:

For 1:

Xbar ± Z s/ sqrt (n)

= 180 ± 1.645 (2 / sqrt (40))

= 180 ± 1.645 (0.316227766)

= 180 ± 0.520194675

= 179.48, 180.52

For 2:

Xbar ± Z s/ sqrt (n)

= 179 ± 1.645 (4 / sqrt (45))

<span>= 179 ± 1.645 (0.596284794)</span>

therefore, the answer is letter b

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A grocery store manager claims that 75% of shoppers purchase bananas as least once a month. Technology was used to simulate choo
Keith_Richards [23]

Answer:

P-value = 0.0333

At 5% level of significance;

0.0333 < 0.05

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Step-by-step explanation:

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To test whether population proportion p is overstated;

Null hypothesis H₀ : p = (75%) = 0.75

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now, sample proportion p" = 64 / 100 = 0.64

from the dot plot below, we will determine the p-value for test { P(p" < 0.64)}

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At 5% level of significance;

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We conclude that proportion of shoppers who bought bananas at least once in the past month is overstated

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3 years ago
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