Solve for the two possible values of X:

A small business ships specialty homemade candies to anywhere in the world. Past records indicate that the weight of orders is n

Fantom [35]

Answer:

90% confidence interval for the true mean weight of orders is between a lower limit of 103.8645 grams and an upper limit of 116.1355 grams.

Step-by-step explanation:

Confidence interval for true mean weight is given as sample mean +/- margin of error (E)

sample mean = 110 g

sample sd = 14 g

n = 16

degree of freedom = n - 1 = 16 - 1 = 15

confidence level = 90% = 0.9

significance level = 1 - C = 1 - 0.9 = 0.1 = 10%

critical value (t) corresponding to 15 degrees of freedom and 10% significance level is 1.753

E = t × sample sd/√n = 1.753×14/√16 = 6.1355 g

Lower limit of sample mean = sample mean - E = 110 - 6.1355 = 103.8645 g

Upper limit of sample mean = sample mean + E = 110 + 6.1355 = 116.1355 g

90% confidence interval is (103.8645, 116.1355)

4 0

2 years ago

Read 2 more answers

A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and

mr Goodwill [35]

Answer:

Bias for the estimator = -0.56

Mean Square Error for the estimator = 6.6311

Step-by-step explanation:

Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.

To find - Determine the bias and the mean squared error for this estimator of the mean.

Proof -

Let us denote

X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)

Now,

An estimate of mean, μ is suggested as

$\mu = \frac{3X_{1} + 4X_{2} }{8}$

Now

Bias for the estimator = E(μ bar) - μ

= $E( \frac{3X_{1} + 4X_{2} }{8}) - 4.5$

= $\frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5$

= $\frac{3(4.5) + 4(4.5)}{8} - 4.5$

= $\frac{13.5 + 18}{8} - 4.5$

= $\frac{31.5}{8} - 4.5$

= 3.9375 - 4.5

= - 0.5625 ≈ -0.56

∴ we get

Bias for the estimator = -0.56

Now,

Mean Square Error for the estimator = E[(μ bar - μ)²]

= Var(μ bar) + [Bias(μ bar, μ)]²

= $Var( \frac{3X_{1} + 4X_{2} }{8}) + 0.3136$

= $\frac{1}{64} Var( {3X_{1} + 4X_{2} }) + 0.3136$

= $\frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})] }) + 0.3136$

= $\frac{1}{64} [{3(57.76) + 4(57.76)}] } + 0.3136$

= $\frac{1}{64} [7(57.76)}] } + 0.3136$

= $\frac{1}{64} [404.32] } + 0.3136$

= $6.3175 + 0.3136$

= 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311

6 0

3 years ago

Yall help “How old am i if 200 is reduced by 2 times my age is 16?”

Nikitich [7]

Answer:

Two times of your age = 16 × 2 = 32

So, 200 reduced = 32 - 200 = - 163 years

<h3>PLEASE MARK ME BRAINLIEST</h3>

5 0

2 years ago

A group of hikers walked 8 and 7/10 miles. And then 5 and 1/5 miles. How far did they walk altogether?

Anton [14]

13 and 9/10 miles altogether

5 0

3 years ago

Circle has rad of 1.5m what angle subtended at center of circle by arc of 1m

Mumz [18]

120 degrees or 0.67 radians

5 0

2 years ago