A random sample of 51 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.03 years, with sample standard deviation s = 0.82 years. However, it is thought that the overall population mean age of coyotes is μ = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use α = 0.01.

Yes the sample data   indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years

Step-by-step explanation:

From the question we are told that

   The sample size is  

    The sample mean is  

    The sample standard deviation is  

    The population mean is  

    The level of significance is  

The null hypothesis  is  

The alternative hypothesis is

The critical value of the the level significance obtained from the critical value table for z-value is

 Now the test statistic is mathematically evaluated as

substituting values  

From that calculated and obtained value we see that the critical value of the level of significance is less than the test statistics so we  reject the null hypothesis

Hence there sufficient evidence to proof that the sample data indicates that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years

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step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)