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The diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.
<h3>When do we use two-sample t-test?</h3>The two-sample t-test is used to determine if two population means are equal.
A nutritionist has developed a diet that she claims will help people lose weight. In this,
- Twelve people were randomly selected to try the diet.
- Their weights were recorded prior to beginning the diet and again after 6 months.
Here are the original weights, in pounds, with the weight after 6 months in parentheses.
- Before 192 212 171 215 180 207 165 168 190 184 200 196
- After 183 196 174 211 160 191 162 175 190 179 189 195
The mean of the weights before 6 moths is,
The mean of the weights after 6 months is,
Standard deviation of both the data is 16.9 and 14.7.
1. Null and Alternative Hypotheses.
The following null and alternative hypotheses need to be tested:
This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.
- (2) Rejection Region
Based on the information provided, the significance level is α=0.05 and the degrees of freedom are df = 22. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:
Hence, it is found that the critical value for this right-tailed test is
, for α=0.05 and df=22
The rejection region for this right-tailed test is,
(3) Test Statistics
Since it is assumed that the population variances are equal, the t-statistic is computed as follows:
- (4) Decision about the null hypothesis
Since it is observed that t=0.967≤tc=1.717 it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value is p=0.1721 and since p=0.1721≥0.05p = 0.1721 it is concluded that the null hypothesis is not rejected.
(5) Conclusion
It is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is greater than μ2, at the α=0.05 significance level.
Confidence Interval
The 95% confidence interval is −7.16<μ<19.66
Thus, the diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.
Learn more about the two-sample t-test here;
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Answer:
− 3 y ' ' − 3 y ' + 3 y = 0 : over-damped
− 2 y ' ' − 4 y ' + 1 y = 0 : over-damped
1 y ' ' + 7 y ' + 5 y = 0: over-damped
Step-by-step explanation:
Using the characteristic equation you can express a differential equation of order n as an algebraic equation of degree n:
This differential equation will have a characteristic equation of the form:
Now, you can classify the solution for a differential equation using a simple method. In order to do it, you just need to use the discriminant.
- If the discriminant is greater than zero, the solution is over-damped
- If the discriminant is less than zero, the solution is under-damped
- If the discriminant is equal to zero, the solution is critically damped
So, given the differential equation:
Which has characteristic equation of the form:
The quadratic polynomial of the form:
Has discriminant:
In this case:
So:
In this case:
Therefore the solution is over-damped.
Now, given the differential equation:
Which has characteristic equation of the form:
The quadratic polynomial of the form:
Has discriminant:
In this case:
So:
In this case:
Therefore the solution is over-damped.
Finally, given the differential equation:
Which has characteristic equation of the form:
The quadratic polynomial of the form:
Has discriminant:
In this case:
So:
In this case:
Therefore the solution is over-damped.