Statistical Abstracts (117th edition) reports that the national average amount a single person spends annually for housing is $10,465. A random sample of 20 householders living in the San Francisco Bay area had a sample mean housing cost $14,575 with standard deviation $4,580. Test to see if the mean housing cost in the San Francisco Bay area is higher than the national average. Use a 1%significance level.
What is the level of significance? State the null and the alternate hypotheses
What sampling distribution should I use (z, t, ..)?
What is the value of the sample test statistic?
How can I find (or estimate) the P-value .
State if you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?
State your conclusions in the context of the application.
2007-04-16 14:18:14 · 4 answers · asked by sweetbolbola 2 in Science & Mathematics ➔ Mathematics
level of sig is 1%. Null is Ho:mu=10465 and Alt is H1:mu>10465. Use student's t. t=(14575-10465)/4580/sq rt20...student's t table should help you find the p-value (use one-tailed test). if p-value > .01 then do not reject, otherwise reject.
2007-04-16 14:25:50 · answer #1 · answered by erselius 3 · 0⤊ 0⤋
(In advance - hard to type stats symbols, might be a little hard to read...)
You could do a t-test
T-tests are of the form ((estimate - hypothesized value) / standard error) compare to t-distribution with df = error df.
In this case, you want something like y(bar) - mu(hypothesized) / (standard dev/sqrt(n)), compare to t dist df = n-1. I'm sure whatever textbook you are using has this formula somewhere.
*The reason you'd want to do a t test as opposed to using a normal distribution (z) is becuase you are taking a random sample from a larger population, and the sample size is small.*
This will give you the value of the test statistic, estimating the p-val may be done using a table or various websites (type p-value calculator into Google to check your table answer for example) by comparing your TS to a t-dist df = n-1.
If the p-val is < .01 (1% significance), you may reject the null hypothesis (data is significant).
The conclusion will depend on your p-val, rejecting the null or not. If you can reject the null, this will imply that yes, the data is significantly different from the hypothesized value. If you cannot, this will imply the data is not significantly different from the hypothesized value.
2007-04-16 14:30:32 · answer #2 · answered by themissy11 1 · 0⤊ 0⤋
I'm rusty on this, but I think you need the standard deviation of the mean for your comparison. This would be 4580/19= 240 or so.
Your null hypothesis is that the bay mean cost differs from the national average. Alternate hypotheses could be some other comparison, since they fix the "beta" of the test. You then form the t-statistic which, if I remember, is the differences in estimated means divided by the standard deviation of the mean. This can be done with 20 degrees of freedom. I'm sure the 1% sig. lev. is exceeded. And I would accept the null hypothesis.
2007-04-16 14:30:14 · answer #3 · answered by cattbarf 7 · 0⤊ 0⤋
The null option is the two types are the comparable So the recommend may be at 5% point the recommend is 5.2 the alternative hypothesis may be that the recommend are diverse so recommend<5.2 or recommend>5.2 may be option. yet no you could desire to think of , that the breeder isn't unterested via the undesirable effect so interior the priority you soak up account purely the 2nd hypothesis
2016-12-26 10:41:17 · answer #4 · answered by letitia 3 · 0⤊ 0⤋