Respuesta :
Answer:
[tex]z=\frac{87.8-90.1}{\frac{17.3}{\sqrt{38}}}=-0.820[/tex]
The p value for this case would be given by:
[tex]p_v =2*P(z<-0.820)=0.412[/tex]
Since the p value is very high ant any significance level we will have enough evidence to FAIL to reject the null hypothesis, so then we can conclude that the true mean is not significantly different from 90.1 and the method not shows a significantly effect in the scores.
Step-by-step explanation:
Information provided
[tex]\bar X=87.8[/tex] represent the sample mean for the scores on the standardized test in the general population of 6th graders
[tex]\sigma=17.3[/tex] represent the population standard deviation
[tex]n=38[/tex] sample size
[tex]\mu_o =90.1[/tex] represent the value that we want to verify
z would represent the statistic
[tex]p_v[/tex] represent the p value for the test
System of hypothesis
We want to verify if the new teaching method negatively affects reading comprehension scores, so then the system of hypothesis for this case are:
Null hypothesis:[tex]\mu =90.1[/tex]
Alternative hypothesis:[tex]\mu \neq 90.1[/tex]
Since we know the population deviation the statistic would be:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]z=\frac{87.8-90.1}{\frac{17.3}{\sqrt{38}}}=-0.820[/tex]
The p value for this case would be given by:
[tex]p_v =2*P(z<-0.820)=0.412[/tex]
Since the p value is very high ant any significance level we will have enough evidence to FAIL to reject the null hypothesis, so then we can conclude that the true mean is not significantly different from 90.1 and the method not shows a significantly effect in the scores.
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 90.1
For the alternative hypothesis,
µ < 90.1
Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is
z = (x - µ)/(σ/√n)
Where
x = sample mean
µ = population mean
σ = population standard deviation
n = number of samples
From the information given,
µ = 90.1
x = 87.8
σ = 17.3
n = 38
z = (87.8 - 90.1)/(17.3/√38) = - 0.82
Looking at the normal distribution table, the probability corresponding to the z score is 0.21
Assume significant level = 0.05
Since alpha, 0.05 < than the p value, 0.21, then we would fail to reject the null hypothesis. Therefore, At a 5% level of significance, there is not enough evidence to conclude that the new teaching method negatively affects reading comprehension scores.
