A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected, and their weights at birth are recorded as:
9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds.
If Alpha = 0.05,
1. What is the critical t-value?
2. What is the decision for a statistically significant change in average weights at birth at the 5% level of significance?

Answers

Answer 1

Answer:

Answer:

1. Critical value t=±2.447

2. The null hypothesis is failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the birth weight significantly differs from 6.6 lbs.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the birth weight significantly differs from 6.6 lbs.

Then, the null and alternative hypothesis are:

$H_0: \mu=6.6\n\nH_a:\mu\neq 6.6$

The significance level is 0.05.

The sample has a size n=7.

The sample mean is M=7.56.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1.18.

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

$s_M=(s)/(√(n))=(1.18)/(√(7))=0.446$

Then, we can calculate the t-statistic as:

$t=(M-\mu)/(s/√(n))=(7.56-6.6)/(0.446)=(0.96)/(0.446)=2.152$

The degrees of freedom for this sample size are:

$df=n-1=7-1=6$

For a two-tailed test with 5% level of significance and 6 degrees of freedom, the critical value for t is ±2.447.

As the test statistic t=2.152 is under 2.447 and over -2.447, it falls in the acceptance region, so the effect is not significant. The null hypothesis is failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the birth weight significantly differs from 6.6 lbs.

Sample mean and standard deviation calculations:

$M=(1)/(n)\sum_(i=1)^n\,x_i\n\n\nM=(1)/(7)(9+7.3+6+. . .+6.6)\n\n\nM=(52.9)/(7)\n\n\nM=7.56\n\n\ns=\sqrt{(1)/(n-1)\sum_(i=1)^n\,(x_i-M)^2}\n\n\ns=\sqrt{(1)/(6)((9-7.56)^2+(7.3-7.56)^2+(6-7.56)^2+. . . +(6.6-7.56)^2)}\n\n\ns=\sqrt{(8.32)/(6)}\n\n\ns=√(1.39)=1.18\n\n\n$

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Your dinner in London cost 82 British pounds. How muchwas it in U.S. dollars?

Some one please help me ASAP!!!

Answers

In U.S. dollars, the equivalent money will be $100.74

What is a expression? What is a mathematical equation? What do you mean by domain and range of a function?

A mathematical expression is made up of terms (constants and variables) separated by mathematical operators.

A mathematical equation is used to equate two expressions.

Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis, observations and results of the given problem.

For any function y = f(x), Domain is the set of all possible values of [y] that exists for different values of [x]. Range is the set of all values of [x] for which [y] exists.

Given is your dinner in London cost 82 British pounds

Then in U.S. dollars, the equivalent money will be equivalent to -

x = 82 pounds = 82 x 1.2286 = $100.74

x = $100.74

Therefore, in U.S. dollars, the equivalent money will be $100.74

To solve more questions on Equations, Equation Modelling and Expressions visit the link below -

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the answer is $108.32

Y/9 + 5 = 0 ???????????