A random sample of n1 = 49 measurements from a population with population standard deviation σ1 = 3 had a sample mean of x1 = 12. An independent random sample of n2 = 64 measurements from a second population with population standard deviation σ2 = 4 had a sample mean of x2 = 14. Test the claim that the population means are different. Use level of significance 0.01.What distribution does the sample test statistic follow? Explain.

Answers

Answer 1

Answer:

Answer:

We reject the null hypothesis that the population means are equal and accept the alternative hypothesis that the population means are different.

Step-by-step explanation:

We have large sample sizes $n_(1) = 49$ and $n_(2) = 64$ , the unbiased point estimate for $\mu_(1)-\mu_(2)$ is $\bar{x}_(1) - \bar{x}_(2)$ , i.e., 12-14 = -2.

The standard error is given by $\sqrt{(\sigma^(2)_(1))/(n_(1))+(\sigma^(2)_(2))/(n_(2))}$ , i.e.,

$\sqrt{((3)^(2))/(49)+((4)^(2))/(64)}$ = 0.6585.

We want to test $H_(0): \mu_(1)-\mu_(2) = 0$ vs $H_(1): \mu_(1)-\mu_(2) \neq 0$ (two-tailed alternative). The rejection region is given by RR = {z | z < -2.5758 or z > 2.5758} where -2.5758 and 2.5758 are the 0.5th and 99.5th quantiles of the standard normal distribution respectively. The test statistic is $Z = \frac{\bar{x}_(1) - \bar{x}_(2)-0}{\sqrt{(\sigma^(2)_(1))/(n_(1))+(\sigma^(2)_(2))/(n_(2))}}$ and the observed value is $z_(0) = (-2)/(0.6585) = -3.0372$ . Because -3.0372 fall inside RR, we reject the null hypothesis.

The test statistic follow a standard normal distribution because we are dealing with large sample sizes.

Answer 2

Answer:

Final answer:

In this scenario of comparing two independent samples and given that the sample sizes are large, the sample test statistic follows the Standard Normal distribution or Z-distribution. The Z-test statistic representing the difference in sample means (in units of standard error) is compared with critical values for a two-tailed test at 0.01 significance level to determine if there's sufficient evidence to reject the null hypothesis that the two population means are equal.

Explanation:

The test in your question pertains to a hypothesis testing scenario featuring two independent samples. This scenario typically involves two population means given that population standard deviations are known. The distribution followed by the sample test statistic in such cases is the Standard Normal distribution or Z-distribution, as the sample sizes (n1 = 49, n2 = 64) are sufficiently large. To test the claim that population means are different (at a significance level of 0.01), you'd typically construct a Z-test statistic that represents the difference in sample means (x1 - x2) in units of its standard error. The Z-test statistic is calculated as follows:

$Z = \frac{x_1 - x_2}{\sqrt{(\sigma_1^2)/(n_1) + (\sigma_2^2)/(n_2)}}$

Here, x1 and x2 are the sample means, σ1 and σ2 are the population standard deviations and n1 and n2 are the samples sizes. The resulting Z-score can be compared with critical Z-scores for a two-tailed test at the given level of significance (0.01) to determine whether or not the null hypothesis (two population means are equal) can be rejected.

Learn more about Z-test here:

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Answers

Answer:

a) $P(X=13)=(24C13)(0.4)^(13) (1-0.4)^(24-13)=0.0608$

$P(X=17)=(24C17)(0.4)^(17) (1-0.4)^(24-17)=0.0017$

$P(X=13 U X=17) = 0.0608+0.0017=0.0624$

b) $P(X \geq 17)=0.000427$

c) $P(X \geq 16)= 1-0.000427=0.9996$

d) $E(X)= np = 22*0.4=8.8$

e) $Var(X) = np(1-p)= 22*0.4*(1-0,4)=5.28$

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=24, p=0.4)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^(n-x)$

Where (nCx) means combinatory and it's given by this formula:

$nCx=(n!)/((n-x)! x!)$

3) Part a

$P(X=13)=(22C13)(0.4)^(13) (1-0.4)^(22-13)=0.0336$

$P(X=17)=(22C17)(0.4)^(17) (1-0.4)^(22-17)=0.00035$

$P(X=13 U X=17) = 0.0336+0.00035=0.0340$

4) Part b

$P(X \geq 17)=P(X=17)+P(X=18)+P(X=19)+P(X=20)+P(X=21)+P(X=22)$

$P(X=17)=(22C17)(0.4)^(17) (1-0.4)^(22-17)=0.00035$

$P(X=18)=(22C18)(0.4)^(18) (1-0.4)^(22-18)=0.0000651$

$P(X=19)=(22C19)(0.4)^(19) (1-0.4)^(22-19)=0.00000914$

$P(X=20)=(22C20)(0.4)^(20) (1-0.4)^(22-20)=0.000000914$

$P(X=21)=(22C21)(0.4)^(21) (1-0.4)^(22-21)=5.81x10^(-8)$

$P(X=22)=(22C22)(0.4)^(22) (1-0.4)^(22-22)=1.76x10^(-9)$

$P(X \geq 17)=0.000427$

5) Part c

$P(X \leq 16)=1-P(X>16)=1-P(X\geq 17) = 1-[P(X=17)+P(X=18)+P(X=19)+P(X=20)+P(X=21)+P(X=22)]$

$P(X \geq 16)= 1-0.000427=0.9996$

6) Part d

The expected value is:

$E(X)= np = 22*0.4=8.8$

7) Part e

The variance is given by:

$Var(X) = np(1-p)= 22*0.4*(1-0,4)=5.28$

Which graph represents the function f(x) = |x + 3|?

Answers

Answer:

Answer is B

Step-by-step explanation:

If you are unsure about where to start, you could always plot some numbers down until you see a general pattern.

But a more intuitive way is to determine what happens during each transformation.

A regular y = |x| will have its vertex at the origin, because nothing is changed for a y = |x| graph. We have a ray that is reflected at the origin about the y-axis.

Now, let's explore the different transformations for an absolute value graph by taking a y = |x + h| graph.

What happens to the graph?

Well, we have shifted the graph -h units, just like a normal trigonometric, linear, or even parabolic graph. That is, we have shifted the graph h units to its negative side (to the left).

What about the y = |x| + h graph?

Well, like a parabola, we shift it h units upwards, and if h is negative, we shift it h units downwards.

So, if you understand what each transformation does, then you would be able to identify the changes in the shape's location.

Answer:

answer is b

Step-by-step explanation:

because i got it right on E D G E

The amount of chlorine needed to treat a swimming pool is directly proportional to the volume of the poolWhat is the constant of proportionality for this relationship

Answers

The amount of chlorine needed to treat a swimming pool is directly proportional to the volume of the poolWhat is the constant of proportionality for this relationship

The answer to this question is 0.002

A certain test preparation course is designed to help students improve their scores on the MCAT exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 6 students' scores on the exam after completing the course: 19, 15, 9, 17, 7, 16. Using these data, construct a 95% confidence interval for the average net change in a students's score after completing the course. Assume the population is approx. normal.Step 1. calculate the sample mean for the given sample data (round answer to 1 decimal place)Step 2. Calculate the sample standard deviation for the given sample data ( round answer to 1 decimal place)Step 3. Find the critical value that should be used in constructing the confidence interval. ( round answer to 3 decimal placed)Step 4. construct the 95% confidence interval (round answer to 1 decimal place)

Answers

Answer:

1) $\bar X =(\sum_(i=1)^n X_i)/(n)=13.8$

2) $s=\sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}=4.8$

3) $t_(\alpha/2)=2.571$

4) The 95% confidence interval would be given by (8.8;18.8)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We can calculate the sample mean and sample deviation with the following formulas:

Step 1. calculate the sample mean for the given sample data (round answer to 1 decimal place)

$\bar X =(\sum_(i=1)^n X_i)/(n)=13.8$

Step 2. Calculate the sample standard deviation for the given sample data ( round answer to 1 decimal place)

$s=\sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}=4.8$

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n=6 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_(\alpha/2)(s)/(√(n))$ (1)

Step 3. Find the critical value that should be used in constructing the confidence interval. ( round answer to 3 decimal places)

In order to calculate the critical value $t_(\alpha/2)$ we need to find first the degrees of freedom, given by:

$df=n-1=6-1=5$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,5)".And we see that $t_(\alpha/2)=2.571$