Question 25
2 answers:
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The coordinates of the reflected point that is reflected across the x-axis to obtain a new point is (1,5).
Answer:
The test statistic t is t=2.9037.
The null hypothesis is rejected.
For a significance level of 0.05, there is enough evidence to support the alternative hypothesis.
Step-by-step explanation:
<em>The question is incomplete:</em>
<em>The sample 1, of size n1=25 has a mean of 1.15 and a standard deviation of 0.31. </em>
<em>The sample 2, of size n2=25 has a mean of 0.95 and a standard deviation of 0.15. </em>
This is a hypothesis test for the difference between populations means.
The null and alternative hypothesis are:
The significance level is α=0.05.
The difference between sample means is Md=0.2.
The estimated standard error of the difference between means is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this test are:
This test is a two-tailed test, with 48 degrees of freedom and t=2.9037, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.0056) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
Answer:
a. 0.313% (0.003134842261), b. 237.79 days (237.788095878)
Step-by-step explanation:
In this case, the length of pregnancies is a normally distributed variable, with a mean of 266 days, and a standard deviation of 15 days.
A graph showing the distribution, with regions of interest for the answer, is presented below.
<h3>First Part: Find the probability of a pregnancy lasting 307 days or longer.</h3>To answer the question regarding <em>the probability of a pregnancy lasting 307 days or longer</em>, it is necessary to calculate what the cumulative probability distribution value is at 307 days. By the way, according to the graph below, 307 days are quite far from the population mean (266 days).
Using the function <em>normaldist(266,15).cdf(307)</em>, from free Desmos software on Internet, we find that, at this length (307 days), the sum of all probabilities for all cases at this value is 99.69% (0.996865157739).
Considering that the total area of the curve is 1, then <em>the probability of pregnancy lasting 307 days or longer</em> is 1 - 0.996865157739 or 0.003134842261 (or 0.00313), approximately 0.313%, a very low probabilty.
This probability is showed as the "light blue" region at the right extreme of the graph.
<h3>Second Part: Find the length that separates premature babies from those who are not premature.</h3>To find the length that separates premature babies from those who are not premature, it is a question about <em>find the days related with the probability of 3% (or 0.03)</em> to find such premature babies. So, it is a question of finding a percentile (or 100-quantiles): given the cumulative normal distribution curve, what is the value (length of pregnancies) that represents this 3%.
Using the function <em>quantile(normaldist(266,15), 0.03)</em>, from free Desmos software on Internet, we obtained a value of 237.79 days (237.788095878) for the length of pregnacies of premature babies. In other words, those babies whose mothers have a length of pregnancy <em>lower</em> than 237.79 days are considered premature, or this is "the length that separates premature babies from those who are not premature".
The area below 237.79 days is the blue shaded region in the graph below, at the left extreme of it.
