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dybincka [34]
3 years ago
14

What is the range of the graphed function?

Mathematics
1 answer:
s2008m [1.1K]3 years ago
5 0

Answer:

{y| y is a real number}

Step-by-step explanation:

The range is all the y's that an equation can produce or all the y's on a graph of the function.

This graph has every and any real y (because it keeps going forever, even off the screen or paper) It's a little hard to see, bc we just see it going left and right mostly, but it is also slightly going up (to the left) and down (to the right)

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Nonpregnant women have an average fasting blood glucose level of about 80 mg/100mL of blood. Researchers want to determine if pr
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Answer:

a) The 90% confidence interval would be given by (71.689;78.311)  

The 95% confidence interval would be given by (71.012;78.988)  

b) p_v =2*P(t_{24}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.  So we have a significant effect.

Step-by-step explanation:

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".  

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".

\bar X=75 represent the sample mean  

\mu population mean (variable of interest)  

s=9.68 represent the sample standard deviation  

n=25 represent the sample size  

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma=0.15

The sample mean \bar X is distributed on this way:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})  

Part a

The confidence interval on this case is given by:

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

We can find the degrees of freedom like this:

df=n-1=25-1=24

confidence 90%

The next step would be find the value of \t_{\alpha/2}, \alpha=1-0.90=0.1 and \alpha/2=0.05  

Using the t distribution with 24 df, excel or a calculator we see that:  

t_{\alpha/2}=1.71

Since we have all the values we can replace:

75 - 1.71\frac{9.68}{\sqrt{25}}=71.689  

75 + 1.71\frac{9.68}{\sqrt{25}}=78.311  

So on this case the 90% confidence interval would be given by (71.689;78.311)  

Confidence 95%

The next step would be find the value of \t_{\alpha/2}, \alpha=1-0.95=0.05 and \alpha/2=0.025  

Using the t distribution with 24 df, excel or a calculator we see that:  

t_{\alpha/2}=2.06

Since we have all the values we can replace:

75 - 2.06\frac{9.68}{\sqrt{25}}=71.012  

75 + 2.06\frac{9.68}{\sqrt{25}}=78.988  

So on this case the 95% confidence interval would be given by (71.012;78.988)  

Part b

Null hypothesis:\mu = 80  

Alternative hypothesis:\mu \neq 80  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{75-80}{\frac{9.68}{\sqrt{25}}}=-2.583  

P-value  

The degrees of freedom are 25-1=24

Since is a two tailed test the p value would given by:  

p_v =2*P(t_{24}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.  So we have a significant effect.

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