When determining how well an observed set of frequencies fits an expected set of frequencies, what is the test statistic

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

6 0

Answer: $\chi^2$ test statistic.

Step-by-step explanation:

The chi-square ( $\chi^2$ ) test is a non-parametric statistical test(also known as Goodness of fit test) which is used to determine if a distribution of observed frequencies differs from the expected frequencies.

Chi-square statistic uses either nominal (categorical) or ordinal level data.

Hence, When determining how well an observed set of frequencies fits an expected set of frequencies, the test statistic is $\chi^2$ test statistic.

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How do measurements of time differ for events in a frame of reference that moves at 50% of the speed of light relative to us? At

ZanzabumX [31]

Answer with explanation:

Relation between Speed , Distance and time

Distance =Speed × Time

→It means Speed is inversely Proportional to time.

As distance will remain constant , in that frame of reference

If speed of light in a Medium

$=s=3 \times 10^8 \frac{\text{meter}}{\text{second}}$

Then, time taken to cross the medium = t seconds or hours or another unit of time.

Now, If speed of light is 50% of the speed of light relative to us

That is Speed of light in another medium

$=w=\frac{50}{100} \times 3 \times 10^8\\\\=1.5 \times 10^8 \frac{\text{meter}}{\text{second}}$

Then, time taken to cross the medium = 2t seconds or hours or another unit of time.

Using Unitary Method

$\rightarrow v_{1}\times t_{1}=v_{2} \times t_{2}\\\\1.\rightarrow 3 \times 10^8 \times t=\frac{50}{100} \times 3 \times 10^8 \times t_{1}\\\\t_{1}=2t\\\\2.\rightarrow 3 \times 10^8 \times t=\frac{99.5}{100} \times 3 \times 10^8 \times t_{2}\\\\t_{2}=\frac{1000t}{995}\\\\t_{2}=\frac{200t}{199}$