site/ region/ rain /temp /plantcov / antspecies/ diversity 2 Dry Steppe 196 5.7

Question

site/ region/ rain /temp /plantcov / antspecies/ diversity

2   Dry Steppe   196   5.7   52   3   0.83
3   Dry Steppe   179   7.0   40   52   1.31
4   Dry Steppe   197   8.0   43   7   1.48
5   Dry Steppe   149   8.5   27   5   0.97
6   Gobi Desert   112   10.7   30   49   0.46
7   Gobi Desert   125   11.4   16   5   1.23
8   Gobi Desert   99   10.9   30   4   *
9   Gobi Desert   125   11.4   56   4   0.76
10   Gobi Desert   84   11.4   22   5   1.26
11   Gobi Desert   115   11.4   14   4   0.69

I have a mini tab project due tomorrow and i dont unerstand how to do this question

Botanist placed seed baits at 11 study sites and observed the number of ants species attracted to each site. the data is listed above. do these data indicated that the average number of any species differs from 5?

a). what are the appropriote null and alternitive hypotheses?

b). do a normal scores plot to check the normal assumption

c). perform the test

what is the value of your test statistics?

what is the p-value of the test?

what is you conclusion using alpha=.05

Explanation / Answer

Here a random variable is antspecies.

a). what are the appropriote null and alternitive hypotheses?

Null hypothesis we denote by H0 and alternative hypothesis we denote by H1.

Here we have to test

H0 : mu = 5 Vs H1 : mu not= 5

where mu is the population mean for antspecies.

Assume alpha = 0.05

Here sample data is given and also population standard deviation is unknown so we use one sample t-test.

Now we can done one sample t-test in MINITAB.

steps :

ENTER data into MINITAB sheet --> Stat --> Basic statistics --> 1-Sample t --> Samples in columns : select data column --> Perform hypothesis test --> Hypothesized mean : 5 --> Options --> Confidence level :95.0 --> Alternative : not equal --> ok --> ok

————— 4/17/2017 11:18:29 AM ————————————————————

Welcome to Minitab, press F1 for help.

One-Sample T: Number of Ant Species

Test of mu = 5 vs not = 5

Variable N Mean StDev SE Mean 95% CI T P
Number of Ant Sp 11 12.8182 18.6752 5.6308 (0.2720, 25.3644) 1.39 0.195

Here test statistic (t) = 1.39

P-value = 0.195

Here P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : There is not sufficient evidence to say that the average number of any species differs from 5

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