Find the percent of change for #1-5

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

4 0

Answer:

1 50.1253%decrease

Step-by-step explanation:

$\frac{399-199}{399}$

= $\frac{100}{399}$

=0.501253

=50.1253

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Find each quotient .<br><br> 3 1/6 ÷ 2

Arlecino [84]

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3 years ago

Here are summary statistics for randomly selected weights of newborn girls: nequals=174174, x overbarxequals=30.930.9 hg, seq

MaRussiya [10]

Answer:

The 95% confidence interval would be given by (29.780;32.020)

Step-by-step explanation:

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".

$\bar X=30.9$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=7.5 represent the sample standard deviation

n=174 represent the sample size

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

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

$df=n-1=174-1=173$

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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,173)".And we see that $t_{\alpha/2}=1.97$ , this value is similar to the obtained with the normal standard distribution since the sample size is large to approximate the t distribution with the normal distribution.