Question

Nonpregnant women have an average fasting blood glucose level of about 80 mg/100mL of blood. Researchers want to determine if pregnancy causes a change in blood glucose level. They take a random sample of 25 third trimester women and find their average fasting blood glucose level is 75 mg/100mL with sample standard deviation of 9.68.(a) Find 90% and 95% confidence intervals for average fasting blood glucose level of third trimester women. (b) Write a suitable nul and alternate hypothesis and then determine if pregnancy has a statistically significant effect on blood glucose level.

Answer 1

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.