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Andrei [34K]
2 years ago
6

Aisha wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 76

cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 8.9 and a standard deviation of 2.3. What is the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Assume the data is from a normally distributed population. Round answers to 3 decimal places where possible.
Mathematics
1 answer:
lozanna [386]2 years ago
4 0

Using the t-distribution, the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies is (8.374, 9.426).

<h3>What is a t-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm t\frac{s}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • t is the critical value.
  • n is the sample size.
  • s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 90% confidence interval, with 76 - 1 = 75 df, is t = 1.9921.

The other parameters are given as follows:

\overline{x} = 8.9, s = 2.3, n = 76.

Hence the bounds of the interval are:

\overline{x} - t\frac{s}{\sqrt{n}} = 8.9 - 1.9921\frac{2.3}{\sqrt{76}} = 8.374

\overline{x} + t\frac{s}{\sqrt{n}} = 8.9 + 1.9921\frac{2.3}{\sqrt{76}} = 9.426

More can be learned about the t-distribution at brainly.com/question/16162795

#SPJ1

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kamal has 2 liters of soup. he pours it into 4 bowls each bowl gets an equal amount how many milliliters of soup are in each bow
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6 0
3 years ago
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According to the last census (2010), the mean number of people per household in the United States is LaTeX: \mu = 2.58 Assume a
Veseljchak [2.6K]

Answer:

P(2.50 < Xbar < 2.66) = 0.046

Step-by-step explanation:

We are given that Population Mean, \mu = 2.58 and Standard deviation, \sigma = 0.75

Also, a random sample (n) of 110 households is taken.

Let Xbar = sample mean household size

The z score probability distribution for sample mean is give by;

             Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)

P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar \leq 2.50)

P(Xbar < 2.66) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{2.66-2.78}{\frac{0.75}{\sqrt{110} } } ) = P(Z < -1.68) = 1 - P(Z  1.68)

                                                              = 1 - 0.95352 = 0.04648

P(Xbar \leq 2.50) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{2.50-2.78}{\frac{0.75}{\sqrt{110} } }  ) = P(Z \leq  -3.92) = 1 - P(Z < 3.92)

                                                              = 1 - 0.99996 = 0.00004  

Therefore, P(2.50 < Xbar < 2.66) = 0.04648 - 0.00004 = 0.046

7 0
3 years ago
A recipe calls for 2 1/2 cups of sugar. You have 2 2/3 cups of sugar. Do you have enough sugar? If not, how much more sugar is n
Karo-lina-s [1.5K]

The difference between the cups of sugar is 1/6, hence you have enough sugar.

<h3>Difference and sum of fractions</h3>

Fractions are written as a ratio of two integers, For instance, a/b is a fraction where a and b are integers.

According to the question, a recipe calls for 2 1/2 cups of sugar. If have 2 2/3 cups of sugar;

Difference of cups of sugar = 2 2/3 - 2 1/2

Convert to improper fraction

Difference = 8/3 -  5/2

Difference = 16-15/6

Difference = 1/6

Since the difference between the cups of sugar is 1/6, hence you have enough sugar.

Learn more on fraction here: brainly.com/question/17220365

#SPJ1

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