drawer contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted. If one item is chosen at random, what is the probability that it is rusted or a bolt ? In a study to test whether there is a difference between the average height of adult females born in two different
countries, random samples yielded the following results :
n1 = 12O, X1 = 62.7 , S1 = 2.50
n2: 150, X2 : 61.8 , S2 = 2.62
where the measurements are in inches. Use 0.05 level of significance to test the difference between ti-re average heights.
First question: one-half = 50%
"X" is typically used to stand for a raw score. If you don't have the symbol, you can write out "mean."
Z = (μ1 - μ2)/Standard error of difference between means (SEμ-μ)
SEμ-μ = √(SEμ1^2 + SEμ2^2)
SEμ = SD/√(n-1)
You are looking for z > 1.96 to reject Ho at .05 level.
I'll let you do the calculations.
To solve the first problem, we need to determine the number of rusted items and the number of bolts in the drawer.
Given that the drawer contains 50 bolts and 150 nuts and that half of the bolts and half of the nuts are rusted, we can calculate the number of rusted items as follows:
Number of rusted bolts = (1/2) * 50 = 25 bolts
Number of rusted nuts = (1/2) * 150 = 75 nuts
Now, to find the probability of selecting an item that is either rusted or a bolt, we need to find the total number of rusted items and bolts and divide it by the total number of items in the drawer.
Total number of rusted items = Number of rusted bolts + Number of rusted nuts = 25 + 75 = 100
Total number of items in the drawer = Number of bolts + Number of nuts = 50 + 150 = 200
Therefore, the probability of selecting an item that is rusted or a bolt is:
Probability = Total number of rusted items / Total number of items in the drawer
= 100 / 200
= 0.5
So, the probability that the randomly chosen item is rusted or a bolt is 0.5 or 50%.
Moving on to the second problem, we are given two samples of heights from two different countries. To test if there is a significant difference between the average heights, we can use a t-test. Here's how we can perform the calculation:
Step 1: State the null and alternative hypotheses:
Null hypothesis (H0): There is no significant difference between the average heights of adult females born in the two different countries.
Alternative hypothesis (Ha): There is a significant difference between the average heights of adult females born in the two different countries.
Step 2: Set the significance level:
In this case, the significance level is given as 0.05, which means we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis when it is true.
Step 3: Calculate the test statistic:
We can use the t-test formula to calculate the test statistic:
t = (X1 - X2) / sqrt((S1^2/n1) + (S2^2/n2))
In this case:
X1 = 62.7 (average height of females in country 1)
X2 = 61.8 (average height of females in country 2)
S1 = 2.50 (standard deviation of heights in country 1)
S2 = 2.62 (standard deviation of heights in country 2)
n1 = 120 (sample size for country 1)
n2 = 150 (sample size for country 2)
Plugging in these values, we can calculate the test statistic.
Step 4: Determine the critical value and make a decision:
Using a t-table or a statistical software, we can find the critical value corresponding to a 0.05 significance level and degrees of freedom equal to the smaller of (n1 - 1) and (n2 - 1).
Step 5: Compare the test statistic with the critical value:
If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
By following these steps, we can determine whether there is a significant difference between the average heights of females born in the two different countries at the 0.05 level of significance.