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Jobisdone [24]
4 years ago
9

SampleVarianceandStandardError: You encounter a deck of Martian playing cards.There are 87 cards in the deck. You cannot read Ma

rtian, and so the meaning of the cards is mysterious. However, you notice that some cards are blue, and others are yellow.(a) You shuffle the deck, and draw one card. You repeat this exercise 10 times, replacing the card you drew each time before shuffling. You see 7 yellow and 3 blue cards in the deck. As you know, the maximum likelihood estimate of the fraction of blue cards in the deck is 0.3. What is the standard error of this estimate?(b) How many times would you need to repeat the exercise to reduce the standard error to 0.05?
Mathematics
1 answer:
docker41 [41]4 years ago
4 0

Answer:

(a) The standard error of estimate of the fraction of blue cards in the deck is 0.145.

(b) The sample size required is 84.

Step-by-step explanation:

The proportion of blue cards is, <em>p</em> = 0.30.

The number of times the cards were shuffled is, <em>n</em> = 10.

(a)

The standard error for sample proportion is given by:

SE_{p}=\sqrt{\frac{p(1-p)}{n}}

Compute the value of standard error as follows:

SE_{p}=\sqrt{\frac{p(1-p)}{n}}\\=\sqrt{\frac{0.30\times (1-0.30)}{10}}\\=\sqrt{0.021}\\=0.145

Thus, the standard error of estimate of the fraction of blue cards in the deck is 0.145.

(b)

Now we need to reduce the standard error.

The standard error is inversely proportional to the sample size.

If the sample size is increased then the standard error will be decreased.

So we need to use a larger sample size.

Compute the value of <em>n</em> for standard error 0.05 and <em>p</em> = 0.30 as follows:

SE_{p}=\sqrt{\frac{p(1-p)}{n}}\\0.05=\sqrt{\frac{0.30(1-0.30)}{n}}\\0.05^{2}=\frac{0.21}{n}\\n=84

Thus, the sample size required is 84.

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Inverse of y equals 12 to the x
e-lub [12.9K]

Answer:

f^{-1}(x) = log_{12}(y)

Step-by-step explanation:

We have the following function

y = 12^x, and we need to find the inverse function.

To find the inverse function we should solve the equation for "x". To do so, first, we need to:

1. Take the logarithm in both sides of the equation:

lg_12 (y) = log _12 (12^x)

(Please read lg_12 as: "Logarithm with base 12")

From property of logarithm, we know that lg (a^b) = b*log(a)

Then:

lg_12 (y) = x*log _12 (12)

We also know that log _12 (12) = 1

Then:

x = log_12(y).

Then, the inverse of: y= 12^x is:

f^{-1}(x) = log_{12}(y)

5 0
3 years ago
My favorite pizza restaurant advertises that their average delivery time is 20 minutes. It always feels like it takes forever fo
andrezito [222]

Answer:

Population parameter(s): mean μ=20

Sample statistics: M=20.7, s=2.1

Hypotheses:

H_0: \mu=20\\\\H_a:\mu> 20

Test Statistic: t=2.357

p-value: 0.011

Reject H? YES

Conclusion (in context of the problem): There is  enough evidence to support the claim that the advertised delivery time is overly optimistic and it is larger than 20 minutes.  

<em>Is this difference practically significant? </em>No, the difference as the sample mean delivery time is under one minute of difference from the advertised time. Although it is significantly from the statistical point of view, the  difference is under 5% of the advertised delivery time.

The 99% confidence interval for the mean is (19.904, 21.496).

With this level of confidence, the null hypothesis failed to be rejected, as t=20 is a possible value for the true mean (is included in the confidence interval).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the advertised delivery time is overly optimistic and it is larger than 20 minutes.  

Then, the null and alternative hypothesis are:

H_0: \mu=20\\\\H_a:\mu> 20

The significance level is 0.05.

The sample has a size n=50.

The sample mean is M=20.7.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.1.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2.1}{\sqrt{50}}=0.297

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{20.7-20}{0.297}=\dfrac{0.7}{0.297}=2.357

The degrees of freedom for this sample size are:

df=n-1=50-1=49

This test is a right-tailed test, with 49 degrees of freedom and t=2.357, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>2.357)=0.011

As the P-value (0.011) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that the advertised delivery time is overly optimistic and it is larger than 20 minutes.  

2. We have to calculate a 99% confidence interval for the mean.

The sample mean is M=20.7.

The sample size is N=50.

The t-value for a 99% confidence interval is t=2.68.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.68 \cdot 0.297=0.796

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 20.7-0.796=19.904\\\\UL=M+t \cdot s_M = 20.7+0.796=21.496

The  99% confidence interval for the mean is (19.904, 21.496).

With this level of confidence, the null hypothesis failed to be rejected, as t=20 is a possible value for the true mean (is included in the confidence interval).

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